Operations on Real Numbers Questions in English

(B) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}-\sqrt{2})$.
$\frac{1}{\sqrt{3}+\sqrt{2}} = \frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}$
$= \frac{\sqrt{3}-\sqrt{2}}{3-2}$
$= \frac{\sqrt{3}-\sqrt{2}}{1} = \sqrt{3}-\sqrt{2}$
Now,substitute the given values $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$:
$= 1.732 - 1.414$
$= 0.318$

(B) Given $a = 5 + 2\sqrt{6}$.
Since $b = \frac{1}{a}$,we rationalize the denominator:
$b = \frac{1}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{5 - 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} = \frac{5 - 2\sqrt{6}}{25 - 24} = 5 - 2\sqrt{6}$.
We need to find $a^2 + b^2$. Using the identity $a^2 + b^2 = (a + b)^2 - 2ab$:
First,calculate $a + b = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10$.
Next,calculate $ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1$.
Substituting these values into the identity:
$a^2 + b^2 = (10)^2 - 2(1) = 100 - 2 = 98$.

(D) To simplify the expression,we rationalize the denominator for each term:
$\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}} \times \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}} = \frac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{10-3} = \frac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{7} = \sqrt{30}-3$
$\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}} \times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} = \frac{2 \sqrt{5}(\sqrt{6}-\sqrt{5})}{6-5} = 2 \sqrt{30}-10$
$\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}} \times \frac{\sqrt{15}-3 \sqrt{2}}{\sqrt{15}-3 \sqrt{2}} = \frac{3 \sqrt{2}(\sqrt{15}-3 \sqrt{2})}{15-18} = \frac{3 \sqrt{30}-18}{-3} = -\sqrt{30}+6$
Now,substitute these back into the original expression:
$(\sqrt{30}-3) - (2 \sqrt{30}-10) - (-\sqrt{30}+6)$
$= \sqrt{30}-3-2 \sqrt{30}+10+\sqrt{30}-6$
$= (\sqrt{30}-2 \sqrt{30}+\sqrt{30}) + (-3+10-6)$
$= 0 + 1 = 1$

(A) Given expression: $\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$
Taking the common denominator $(3\sqrt{3}-2\sqrt{2})(3\sqrt{3}+2\sqrt{2})$:
$= \frac{4(3\sqrt{3}+2\sqrt{2}) + 3(3\sqrt{3}-2\sqrt{2})}{(3\sqrt{3})^2 - (2\sqrt{2})^2}$
$= \frac{12\sqrt{3} + 8\sqrt{2} + 9\sqrt{3} - 6\sqrt{2}}{27 - 8}$
$= \frac{21\sqrt{3} + 2\sqrt{2}}{19}$
Substituting the values $\sqrt{3}=1.732$ and $\sqrt{2}=1.414$:
$= \frac{21(1.732) + 2(1.414)}{19}$
$= \frac{36.372 + 2.828}{19}$
$= \frac{39.2}{19} \approx 2.063$

(B) Given,$a = \frac{3+\sqrt{5}}{2}$.
First,calculate $a^{2}$:
$a^{2} = \left(\frac{3+\sqrt{5}}{2}\right)^{2} = \frac{9 + 5 + 6\sqrt{5}}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2}$.
Next,calculate $\frac{1}{a^{2}}$ by rationalizing the denominator:
$\frac{1}{a^{2}} = \frac{2}{7 + 3\sqrt{5}} = \frac{2(7 - 3\sqrt{5})}{(7 + 3\sqrt{5})(7 - 3\sqrt{5})} = \frac{2(7 - 3\sqrt{5})}{49 - 45} = \frac{2(7 - 3\sqrt{5})}{4} = \frac{7 - 3\sqrt{5}}{2}$.
Finally,add $a^{2}$ and $\frac{1}{a^{2}}$:
$a^{2} + \frac{1}{a^{2}} = \frac{7 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2} = \frac{7 + 3\sqrt{5} + 7 - 3\sqrt{5}}{2} = \frac{14}{2} = 7$.

(C) Rationalizing the denominator for $x$:
$x = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{(\sqrt{3}+\sqrt{2})^{2}}{3-2} = 3+2+2\sqrt{6} = 5+2\sqrt{6}$.
Similarly,for $y$:
$y = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2} = 3+2-2\sqrt{6} = 5-2\sqrt{6}$.
Now,calculate $x+y$ and $xy$:
$x+y = (5+2\sqrt{6}) + (5-2\sqrt{6}) = 10$.
$xy = (5+2\sqrt{6})(5-2\sqrt{6}) = 5^{2} - (2\sqrt{6})^{2} = 25 - 24 = 1$.
Using the identity $x^{2}+y^{2} = (x+y)^{2} - 2xy$:
$x^{2}+y^{2} = (10)^{2} - 2(1) = 100 - 2 = 98$.

(N/A) To find rational numbers between $\frac{2}{9}$ and $\frac{2}{7}$,we first make their denominators equal by finding the least common multiple of $9$ and $7$,which is $63$.
$\frac{2}{9} = \frac{2 \times 7}{9 \times 7} = \frac{14}{63}$
$\frac{2}{7} = \frac{2 \times 9}{7 \times 9} = \frac{18}{63}$
Since there are only three integers between $14$ and $18$ $(15, 16, 17)$,we multiply the numerator and denominator by a larger factor,say $2$,to get more space.
$\frac{14}{63} = \frac{14 \times 2}{63 \times 2} = \frac{28}{126}$
$\frac{18}{63} = \frac{18 \times 2}{63 \times 2} = \frac{36}{126}$
Now,we can choose any four rational numbers between $\frac{28}{126}$ and $\frac{36}{126}$,such as $\frac{29}{126}, \frac{30}{126}, \frac{31}{126}, \text{ and } \frac{32}{126}$.
Simplifying these,we get $\frac{29}{126}, \frac{5}{21}, \frac{31}{126}, \text{ and } \frac{16}{63}$.

(A) To find three rational numbers between $\frac{2}{3}$ and $\frac{4}{5}$,first make the denominators equal by finding the least common multiple $(LCM)$ of $3$ and $5$,which is $15$.
Convert the fractions: $\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$ and $\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}$.
Since we need three numbers,multiply the numerator and denominator of both fractions by $(3 + 1) = 4$ (or any number greater than $3$).
$\frac{10}{15} = \frac{10 \times 4}{15 \times 4} = \frac{40}{60}$ and $\frac{12}{15} = \frac{12 \times 4}{15 \times 4} = \frac{48}{60}$.
Three rational numbers between $\frac{40}{60}$ and $\frac{48}{60}$ are $\frac{41}{60}, \frac{42}{60}, \text{and } \frac{43}{60}$.

(B) To find three rational numbers between $\frac{1}{7}$ and $\frac{3}{7}$,we can multiply the numerator and denominator of both fractions by $(3 + 1) = 4$ or any larger integer to increase the gap.
Multiplying by $2$ (since $2$ is between $1$ and $3$):
$\frac{1}{7} = \frac{1 \times 2}{7 \times 2} = \frac{2}{14}$
$\frac{3}{7} = \frac{3 \times 2}{7 \times 2} = \frac{6}{14}$
The rational numbers between $\frac{2}{14}$ and $\frac{6}{14}$ are $\frac{3}{14}, \frac{4}{14}, \text{and } \frac{5}{14}$.
Thus,the three rational numbers are $\frac{3}{14}, \frac{2}{7}, \text{and } \frac{5}{14}$.

To find five rational numbers between $\frac{2}{7}$ and $\frac{2}{5}$,we first make the denominators the same by finding the least common multiple $(LCM)$ of $7$ and $5$,which is $35$.
$\frac{2}{7} = \frac{2 \times 5}{7 \times 5} = \frac{10}{35}$
$\frac{2}{5} = \frac{2 \times 7}{5 \times 7} = \frac{14}{35}$
Since we need five rational numbers,we can multiply the numerator and denominator of both fractions by $6$ (or any number greater than $5$):
$\frac{10 \times 6}{35 \times 6} = \frac{60}{210}$
$\frac{14 \times 6}{35 \times 6} = \frac{84}{210}$
Now,we can choose any five rational numbers between $\frac{60}{210}$ and $\frac{84}{210}$,such as $\frac{61}{210}, \frac{62}{210}, \frac{63}{210}, \frac{64}{210}, \text{ and } \frac{65}{210}$.

(B) To find five rational numbers between $-\frac{2}{3}$ and $\frac{1}{5}$,first make the denominators equal by finding the least common multiple $(LCM)$ of $3$ and $5$,which is $15$.
$-\frac{2}{3} = -\frac{2 \times 5}{3 \times 5} = -\frac{10}{15}$
$\frac{1}{5} = \frac{1 \times 3}{5 \times 3} = \frac{3}{15}$
Now,we need to find five rational numbers between $-\frac{10}{15}$ and $\frac{3}{15}$.
The integers between $-10$ and $3$ are $-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2$.
We can choose any five from these,for example: $-\frac{9}{15}, -\frac{8}{15}, -\frac{7}{15}, -\frac{6}{15}, -\frac{5}{15}$.
Simplifying these,we get: $-\frac{3}{5}, -\frac{8}{15}, -\frac{7}{15}, -\frac{2}{5}, -\frac{1}{3}$.

(TRUE) The statement is True.
$A$ whole number is any number belonging to the set $W = \{0, 1, 2, 3, ...\}$.
$A$ rational number is defined as any number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Any whole number $n$ can be written as $\frac{n}{1}$.
For example,$0 = \frac{0}{1}$,$1 = \frac{1}{1}$,$2 = \frac{2}{1}$,etc.
Since every whole number can be expressed as a ratio of two integers with a non-zero denominator,every whole number is a rational number.

(A) To add the given expressions,we group the like terms involving $\sqrt{3}$ and $\sqrt{5}$ together.
$(4 \sqrt{3}+2 \sqrt{5})+(6 \sqrt{3}-4 \sqrt{5})$
$= (4 \sqrt{3}+6 \sqrt{3}) + (2 \sqrt{5}-4 \sqrt{5})$
$= (4+6) \sqrt{3} + (2-4) \sqrt{5}$
$= 10 \sqrt{3} - 2 \sqrt{5}$

(B) To multiply $3 \sqrt{7}$ and $5 \sqrt{7}$,we follow these steps:
Step $1$: Multiply the coefficients (the numbers outside the square roots) together: $3 \times 5 = 15$.
Step $2$: Multiply the square root terms together: $\sqrt{7} \times \sqrt{7} = 7$.
Step $3$: Multiply the results from Step $1$ and Step $2$: $15 \times 7 = 105$.
Therefore,$3 \sqrt{7} \times 5 \sqrt{7} = 105$.

(A) To divide $12 \sqrt{30}$ by $3 \sqrt{5}$,we write it as a fraction:
$\frac{12 \sqrt{30}}{3 \sqrt{5}}$
We can simplify the coefficients: $\frac{12}{3} = 4$.
Next,we simplify the square roots using the property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$:
$\sqrt{\frac{30}{5}} = \sqrt{6}$
Combining these,we get $4 \times \sqrt{6} = 4 \sqrt{6}$.

(A) To simplify the expression $(3+\sqrt{5})(4-\sqrt{11})$,we use the distributive property of multiplication over addition ($FOIL$ method):
$(3+\sqrt{5})(4-\sqrt{11}) = 3(4-\sqrt{11}) + \sqrt{5}(4-\sqrt{11})$
$= (3 \times 4) - (3 \times \sqrt{11}) + (\sqrt{5} \times 4) - (\sqrt{5} \times \sqrt{11})$
$= 12 - 3\sqrt{11} + 4\sqrt{5} - \sqrt{5 \times 11}$
$= 12 - 3\sqrt{11} + 4\sqrt{5} - \sqrt{55}$

(A) To simplify the expression $(\sqrt{15}+\sqrt{7})(\sqrt{15}-\sqrt{7})$,we use the algebraic identity $(a+b)(a-b) = a^2 - b^2$.
Here,$a = \sqrt{15}$ and $b = \sqrt{7}$.
Applying the identity:
$(\sqrt{15})^2 - (\sqrt{7})^2 = 15 - 7 = 8$.

(A) To simplify the expression $(\sqrt{11}-\sqrt{3})^{2}$,we use the algebraic identity $(a-b)^{2} = a^{2} - 2ab + b^{2}$.
Here,$a = \sqrt{11}$ and $b = \sqrt{3}$.
Substituting these values into the identity:
$(\sqrt{11}-\sqrt{3})^{2} = (\sqrt{11})^{2} - 2(\sqrt{11})(\sqrt{3}) + (\sqrt{3})^{2}$
$= 11 - 2\sqrt{11 \times 3} + 3$
$= 11 - 2\sqrt{33} + 3$
$= 14 - 2\sqrt{33}$

(N/A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(5 \sqrt{3} + 3 \sqrt{5})$.
$\frac{30}{5 \sqrt{3}-3 \sqrt{5}} = \frac{30}{5 \sqrt{3}-3 \sqrt{5}} \times \frac{5 \sqrt{3}+3 \sqrt{5}}{5 \sqrt{3}+3 \sqrt{5}}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{(5 \sqrt{3})^{2}-(3 \sqrt{5})^{2}}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{75-45}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{30}$
$= 5 \sqrt{3}+3 \sqrt{5}$

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(5-2 \sqrt{3})$.
$\frac{1}{5+2 \sqrt{3}} = \frac{1}{5+2 \sqrt{3}} \times \frac{5-2 \sqrt{3}}{5-2 \sqrt{3}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{5-2 \sqrt{3}}{(5)^2 - (2 \sqrt{3})^2}$
$= \frac{5-2 \sqrt{3}}{25 - (4 \times 3)}$
$= \frac{5-2 \sqrt{3}}{25 - 12}$
$= \frac{5-2 \sqrt{3}}{13}$

To rationalize the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $\sqrt{m^{2}+n^{2}}-m$.
$\frac{n^{2}}{\sqrt{m^{2}+n^{2}}+m} = \frac{n^{2}}{\sqrt{m^{2}+n^{2}}+m} \times \frac{\sqrt{m^{2}+n^{2}}-m}{\sqrt{m^{2}+n^{2}}-m}$
Using the identity $(a+b)(a-b) = a^{2}-b^{2}$ in the denominator:
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{(\sqrt{m^{2}+n^{2}})^{2} - (m)^{2}}$
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{m^{2}+n^{2}-m^{2}}$
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{n^{2}}$
$= \sqrt{m^{2}+n^{2}}-m$

(A) Given: $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}=a+b \sqrt{30}$
To rationalize the denominator,multiply the numerator and denominator by $(3 \sqrt{5}+2 \sqrt{6})$:
$\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}} \times \frac{3 \sqrt{5}+2 \sqrt{6}}{3 \sqrt{5}+2 \sqrt{6}}=a+b \sqrt{30}$
Using the identity $(x-y)(x+y) = x^2 - y^2$ in the denominator:
$\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5})^{2}-(2 \sqrt{6})^{2}}=a+b \sqrt{30}$
Expanding the numerator:
$(2 \sqrt{6})(3 \sqrt{5}) + (2 \sqrt{6})(2 \sqrt{6}) - (\sqrt{5})(3 \sqrt{5}) - (\sqrt{5})(2 \sqrt{6}) = 6 \sqrt{30} + 24 - 15 - 2 \sqrt{30} = 9 + 4 \sqrt{30}$
Calculating the denominator:
$(3 \sqrt{5})^2 - (2 \sqrt{6})^2 = 45 - 24 = 21$
So,$\frac{9+4 \sqrt{30}}{21} = a+b \sqrt{30}$
$\frac{9}{21} + \frac{4 \sqrt{30}}{21} = a+b \sqrt{30}$
$\frac{3}{7} + \frac{4}{21} \sqrt{30} = a+b \sqrt{30}$
Comparing the rational and irrational parts,we get $a=\frac{3}{7}$ and $b=\frac{4}{21}$.

(N/A) Given: $x = 5 + 2\sqrt{6}$.
First,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{5 + 2\sqrt{6}} = \frac{1}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{5 - 2\sqrt{6}}{25 - 24} = 5 - 2\sqrt{6}$.
Now,calculate $x + \frac{1}{x}$:
$x + \frac{1}{x} = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10$.
To find $x^{2} + \frac{1}{x^{2}}$:
$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2 = (10)^{2} - 2 = 100 - 2 = 98$.
To find $x^{3} + \frac{1}{x^{3}}$:
$x^{3} + \frac{1}{x^{3}} = (x + \frac{1}{x})^{3} - 3(x + \frac{1}{x}) = (10)^{3} - 3(10) = 1000 - 30 = 970$.

(N/A) To add the given expressions,we group the like terms:
$(4 \sqrt{6}-8 \sqrt{10}) + (3 \sqrt{6}+12 \sqrt{10})$
$= (4 \sqrt{6} + 3 \sqrt{6}) + (-8 \sqrt{10} + 12 \sqrt{10})$
$= (4 + 3) \sqrt{6} + (-8 + 12) \sqrt{10}$
$= 7 \sqrt{6} + 4 \sqrt{10}$

(A) To multiply $5 \sqrt{3}$ and $4 \sqrt{12}$,we follow these steps:
$1$. Write the expression as $(5 \sqrt{3}) \times (4 \sqrt{12})$.
$2$. Multiply the coefficients: $5 \times 4 = 20$.
$3$. Multiply the radicals: $\sqrt{3} \times \sqrt{12} = \sqrt{3 \times 12} = \sqrt{36}$.
$4$. Calculate the square root: $\sqrt{36} = 6$.
$5$. Multiply the results: $20 \times 6 = 120$.
Therefore,the product is $120$.

(B) To simplify the expression $5 \sqrt{2} + 2 \sqrt{8} - 3 \sqrt{32} + 4 \sqrt{128}$, we first simplify each radical term:
$1$. $5 \sqrt{2} = 5 \sqrt{2}$
$2$. $2 \sqrt{8} = 2 \sqrt{4 \times 2} = 2 \times 2 \sqrt{2} = 4 \sqrt{2}$
$3$. $3 \sqrt{32} = 3 \sqrt{16 \times 2} = 3 \times 4 \sqrt{2} = 12 \sqrt{2}$
$4$. $4 \sqrt{128} = 4 \sqrt{64 \times 2} = 4 \times 8 \sqrt{2} = 32 \sqrt{2}$
Now, substitute these values back into the expression:
$5 \sqrt{2} + 4 \sqrt{2} - 12 \sqrt{2} + 32 \sqrt{2}$
Combine the coefficients:
$(5 + 4 - 12 + 32) \sqrt{2} = 29 \sqrt{2}$

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{5} + \sqrt{3})$.
$\frac{1}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
$= \frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2}$
$= \frac{\sqrt{5}+\sqrt{3}}{5 - 3}$
$= \frac{\sqrt{5}+\sqrt{3}}{2}$

(A) To rationalise the denominator of $\frac{4}{\sqrt{10}+\sqrt{6}}$,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{10}-\sqrt{6})$.
$\frac{4}{\sqrt{10}+\sqrt{6}} \times \frac{\sqrt{10}-\sqrt{6}}{\sqrt{10}-\sqrt{6}}$
$= \frac{4(\sqrt{10}-\sqrt{6})}{(\sqrt{10})^2 - (\sqrt{6})^2}$
$= \frac{4(\sqrt{10}-\sqrt{6})}{10-6}$
$= \frac{4(\sqrt{10}-\sqrt{6})}{4}$
$= \sqrt{10}-\sqrt{6}$

(N/A) To rationalise the denominator,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(3 \sqrt{2} + 2 \sqrt{3})$.
$\frac{18}{3 \sqrt{2}-2 \sqrt{3}} \times \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^2 - (2 \sqrt{3})^2}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{(9 \times 2) - (4 \times 3)}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{18 - 12}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{6}$
$= 3(3 \sqrt{2}+2 \sqrt{3})$

(A) To rationalise the denominator of $\frac{1}{7-4 \sqrt{3}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $7+4 \sqrt{3}$.
$\frac{1}{7-4 \sqrt{3}} \times \frac{7+4 \sqrt{3}}{7+4 \sqrt{3}} = \frac{7+4 \sqrt{3}}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{7+4 \sqrt{3}}{49 - (16 \times 3)}$
$= \frac{7+4 \sqrt{3}}{49 - 48}$
$= \frac{7+4 \sqrt{3}}{1}$
$= 7+4 \sqrt{3}$

$(17+12 \sqrt{2})$ To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(3+2 \sqrt{2})$.
$\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}} \times \frac{3+2 \sqrt{2}}{3+2 \sqrt{2}}$
$= \frac{(3+2 \sqrt{2})^2}{(3)^2 - (2 \sqrt{2})^2}$
$= \frac{3^2 + (2 \sqrt{2})^2 + 2(3)(2 \sqrt{2})}{9 - (4 \times 2)}$
$= \frac{9 + 8 + 12 \sqrt{2}}{9 - 8}$
$= \frac{17 + 12 \sqrt{2}}{1}$
$= 17 + 12 \sqrt{2}$

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(5-2 \sqrt{6})$.
$\frac{5-2 \sqrt{6}}{5+2 \sqrt{6}} \times \frac{5-2 \sqrt{6}}{5-2 \sqrt{6}}$
$= \frac{(5-2 \sqrt{6})^2}{(5)^2 - (2 \sqrt{6})^2}$
$= \frac{5^2 - 2(5)(2 \sqrt{6}) + (2 \sqrt{6})^2}{25 - (4 \times 6)}$
$= \frac{25 - 20 \sqrt{6} + 24}{25 - 24}$
$= \frac{49 - 20 \sqrt{6}}{1}$
$= 49 - 20 \sqrt{6}$

(A) Given expression is $\frac{\sqrt{5}}{\sqrt{10}}$.
We can simplify this as $\sqrt{\frac{5}{10}} = \sqrt{\frac{1}{2}}$.
This is equal to $\frac{1}{\sqrt{2}}$.
Rationalizing the denominator,we get $\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
Substituting the value of $\sqrt{2} = 1.4142$,we get $\frac{1.4142}{2} = 0.7071$.

(B) Given that $\sqrt{5} = 2.236$.
We need to evaluate $\frac{4 - \sqrt{5}}{\sqrt{5}}$.
This expression can be simplified as $\frac{4}{\sqrt{5}} - 1$.
Substituting the value of $\sqrt{5}$: $\frac{4}{2.236} - 1$.
Calculating $\frac{4}{2.236} \approx 1.788895$.
Subtracting $1$ from this value gives $1.788895 - 1 = 0.788895$.
Rounding to four decimal places,we get $0.7889$.
However,checking the calculation $\frac{4 - 2.236}{2.236} = \frac{1.764}{2.236} \approx 0.788886$.
Rounding to four decimal places,we get $0.7889$.
Given the options provided,$0.7888$ is the closest match.

(N/A) Given $x = 3 + 2\sqrt{2}$.
First,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2}$.
Now,calculate $x + \frac{1}{x}$:
$x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$.
To find $x^{2} + \frac{1}{x^{2}}$,use the identity $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$:
$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2 = (6)^{2} - 2 = 36 - 2 = 34$.
To find $x^{3} + \frac{1}{x^{3}}$,use the identity $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x + \frac{1}{x})$:
$x^{3} + \frac{1}{x^{3}} = (x + \frac{1}{x})^{3} - 3(x + \frac{1}{x}) = (6)^{3} - 3(6) = 216 - 18 = 198$.
Thus,the values are $34$ and $198$.

(D) Given $x = 7 - 4\sqrt{3}$.
First,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}}$.
Rationalizing the denominator:
$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}} \times \frac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{7 + 4\sqrt{3}}{49 - (16 \times 3)} = \frac{7 + 4\sqrt{3}}{49 - 48} = 7 + 4\sqrt{3}$.
Now,find $x + \frac{1}{x}$:
$x + \frac{1}{x} = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) = 14$.
We know that $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$.
Therefore,$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2$.
Substituting the value:
$x^{2} + \frac{1}{x^{2}} = (14)^{2} - 2 = 196 - 2 = 194$.

(A) To solve $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=a+b \sqrt{35}$,we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{7}+\sqrt{5})$.
$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} \times \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} = \frac{(\sqrt{7}+\sqrt{5})^2}{(\sqrt{7})^2-(\sqrt{5})^2}$
$= \frac{7+5+2\sqrt{35}}{7-5}$
$= \frac{12+2\sqrt{35}}{2}$
$= 6+\sqrt{35}$
Comparing $6+\sqrt{35}$ with $a+b\sqrt{35}$,we get $a=6$ and $b=1$.

(N/A) To solve this,we rationalize each term in the expression.
Step $1$: Rationalize the first term: $\frac{1}{1+\sqrt{2}} = \frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$.
Step $2$: Rationalize the second term: $\frac{1}{\sqrt{2}+\sqrt{3}} = \frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3-2} = \sqrt{3}-\sqrt{2}$.
Step $3$: Rationalize the third term: $\frac{1}{\sqrt{3}+\sqrt{4}} = \frac{1}{\sqrt{4}+\sqrt{3}} \times \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}} = \frac{\sqrt{4}-\sqrt{3}}{4-3} = \sqrt{4}-\sqrt{3} = 2-\sqrt{3}$.
Step $4$: Add the results: $(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + (2-\sqrt{3})$.
Step $5$: Simplify the expression: $\sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} = -1 + 2 = 1$.
Thus,the expression equals $1$.

(A) To find the values of $a$ and $b$,we rationalize the denominator of the given expression:
$\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}$
$= \frac{(5+3 \sqrt{3})(7-4 \sqrt{3})}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{35 - 20 \sqrt{3} + 21 \sqrt{3} - 12(3)}{49 - 16(3)}$
$= \frac{35 + \sqrt{3} - 36}{49 - 48}$
$= \frac{-1 + \sqrt{3}}{1}$
$= -1 + 1 \sqrt{3}$
Comparing this with $a+b \sqrt{3}$,we get $a = -1$ and $b = 1$.

(N/A) Given $x = 2 + \sqrt{3}.$
First,find $\frac{1}{x} = \frac{1}{2 + \sqrt{3}}.$
Rationalizing the denominator: $\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.$
Now,$x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4.$
To find $x^{2} + \frac{1}{x^{2}},$ use the identity $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2.$
$4^{2} = x^{2} + \frac{1}{x^{2}} + 2 \implies 16 = x^{2} + \frac{1}{x^{2}} + 2 \implies x^{2} + \frac{1}{x^{2}} = 14.$
To find $x^{3} + \frac{1}{x^{3}},$ use the identity $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x + \frac{1}{x}).$
$4^{3} = x^{3} + \frac{1}{x^{3}} + 3(4) \implies 64 = x^{3} + \frac{1}{x^{3}} + 12 \implies x^{3} + \frac{1}{x^{3}} = 52.$
Thus,the values are $14$ and $52.$

(D) To evaluate $\frac{\sqrt{2}}{2+\sqrt{2}}$,we first rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(2-\sqrt{2})$:
$\frac{\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2} - (\sqrt{2})^2}{2^2 - (\sqrt{2})^2}$
$= \frac{2\sqrt{2} - 2}{4 - 2} = \frac{2(\sqrt{2} - 1)}{2} = \sqrt{2} - 1$
Given $\sqrt{2} = 1.414$,we substitute this value:
$1.414 - 1 = 0.414$
Thus,the correct value is $0.414$.

(A) To find the value of $\frac{1}{\sqrt{3} + \sqrt{2}}$,we first rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(\sqrt{3} - \sqrt{2})$.
$\frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}$
$= \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \frac{\sqrt{3} - \sqrt{2}}{1} = \sqrt{3} - \sqrt{2}$
Given $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$,we substitute these values:
$1.732 - 1.414 = 0.318$
Thus,the value is $0.318$.

(B) Given the expression: $\frac{\sqrt{10} - \sqrt{5}}{2}$
We can rewrite $\sqrt{10}$ as $\sqrt{2} \times \sqrt{5}$.
So, the expression becomes: $\frac{\sqrt{2} \times \sqrt{5} - \sqrt{5}}{2}$
Factor out $\sqrt{5}$ from the numerator: $\frac{\sqrt{5}(\sqrt{2} - 1)}{2}$
Substitute the given values $\sqrt{5} = 2.236$ and $\sqrt{2} = 1.414$:
$= \frac{2.236(1.414 - 1)}{2}$
$= \frac{2.236(0.414)}{2}$
$= 1.118 \times 0.414$
$= 0.462852$
Rounding to three decimal places, we get $0.463$.

(C) Given expression: $\frac{4}{3 \sqrt{3}-2 \sqrt{2}} + \frac{3}{3 \sqrt{3}+2 \sqrt{2}}$
Rationalize the first term: $\frac{4(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3}-2 \sqrt{2})(3 \sqrt{3}+2 \sqrt{2})} = \frac{12 \sqrt{3}+8 \sqrt{2}}{(3 \sqrt{3})^2 - (2 \sqrt{2})^2} = \frac{12 \sqrt{3}+8 \sqrt{2}}{27-8} = \frac{12 \sqrt{3}+8 \sqrt{2}}{19}$
Rationalize the second term: $\frac{3(3 \sqrt{3}-2 \sqrt{2})}{(3 \sqrt{3}+2 \sqrt{2})(3 \sqrt{3}-2 \sqrt{2})} = \frac{9 \sqrt{3}-6 \sqrt{2}}{27-8} = \frac{9 \sqrt{3}-6 \sqrt{2}}{19}$
Adding both terms: $\frac{12 \sqrt{3}+8 \sqrt{2} + 9 \sqrt{3}-6 \sqrt{2}}{19} = \frac{21 \sqrt{3}+2 \sqrt{2}}{19}$
Substitute the values $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$:
$= \frac{21(1.732) + 2(1.414)}{19} = \frac{36.372 + 2.828}{19} = \frac{39.2}{19} \approx 2.06315...$
Rounding to three decimal places,we get $2.063$.

(A) Given that $a = \frac{\sqrt{5}}{8}$.
We are given the equation $\frac{8}{a} = b \sqrt{5}$.
Substitute the value of $a$ into the equation:
$\frac{8}{\frac{\sqrt{5}}{8}} = b \sqrt{5}$
$8 \times \frac{8}{\sqrt{5}} = b \sqrt{5}$
$\frac{64}{\sqrt{5}} = b \sqrt{5}$
To solve for $b$,multiply both sides by $\frac{1}{\sqrt{5}}$:
$b = \frac{64}{\sqrt{5} \times \sqrt{5}}$
$b = \frac{64}{5}$.

(B) The statement is False.
According to the laws of radicals,for any positive real numbers $a$ and $b$,$\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.
Applying this rule: $\sqrt{3} \times \sqrt{5} = \sqrt{3 \times 5} = \sqrt{15}$.
Since $\sqrt{15} \neq \sqrt{8}$,the given statement is false.

(C) To solve the expression $(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})$,we use the algebraic identity $(a+b)(a-b) = a^2 - b^2$.
Here,$a = \sqrt{5}$ and $b = \sqrt{3}$.
Substituting these values into the identity:
$(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2$.
Therefore,the correct option is $C$.

(B) To find the rationalising factor of an expression of the form $a - \sqrt{b}$,we multiply it by its conjugate $a + \sqrt{b}$ to eliminate the square root.
Here,the given expression is $4 - \sqrt{5}$.
Multiplying $4 - \sqrt{5}$ by $4 + \sqrt{5}$ gives $(4)^2 - (\sqrt{5})^2 = 16 - 5 = 11$,which is a rational number.
Therefore,the rationalising factor of $4 - \sqrt{5}$ is $4 + \sqrt{5}$.

(A) Given the expression: $(\sqrt{5}+3)^{2} = a+b\sqrt{5}$.
Using the algebraic identity $(x+y)^{2} = x^{2} + 2xy + y^{2}$,where $x = \sqrt{5}$ and $y = 3$:
$(\sqrt{5})^{2} + 2(\sqrt{5})(3) + (3)^{2} = a+b\sqrt{5}$.
$5 + 6\sqrt{5} + 9 = a+b\sqrt{5}$.
$14 + 6\sqrt{5} = a+b\sqrt{5}$.
By comparing the rational and irrational parts on both sides,we get $a = 14$ and $b = 6$.

(A) The square root of a number $x$ is a value $y$ such that $y^2 = x$.
Since $7^2 = 7 \times 7 = 49$,the principal square root of $49$ is $7$.
Therefore,$\sqrt{49} = 7$.