Operations on Real Numbers Questions in English

(A) To add the given expressions,group the like terms (terms with the same square root factor):
$(2 \sqrt{2} + 5 \sqrt{3}) + (\sqrt{2} - 3 \sqrt{3})$
$= (2 \sqrt{2} + \sqrt{2}) + (5 \sqrt{3} - 3 \sqrt{3})$
$= (2 + 1) \sqrt{2} + (5 - 3) \sqrt{3}$
$= 3 \sqrt{2} + 2 \sqrt{3}$

(B) To multiply $6 \sqrt{5}$ by $2 \sqrt{5}$,we follow these steps:
$1$. Multiply the rational coefficients: $6 \times 2 = 12$.
$2$. Multiply the square root terms: $\sqrt{5} \times \sqrt{5} = 5$.
$3$. Multiply the results together: $12 \times 5 = 60$.
Thus,$6 \sqrt{5} \times 2 \sqrt{5} = 60$.

(C) To divide $8 \sqrt{15}$ by $2 \sqrt{3}$,we write it as a fraction:
$\frac{8 \sqrt{15}}{2 \sqrt{3}}$
We can simplify the expression by splitting $\sqrt{15}$ into $\sqrt{3} \times \sqrt{5}$:
$= \frac{8 \times \sqrt{3} \times \sqrt{5}}{2 \sqrt{3}}$
Cancel out the common term $\sqrt{3}$ from the numerator and the denominator:
$= \frac{8}{2} \times \sqrt{5}$
$= 4 \sqrt{5}$

$(i)$ Using the distributive property: $(5+\sqrt{7})(2+\sqrt{5}) = 5(2) + 5(\sqrt{5}) + \sqrt{7}(2) + \sqrt{7}(\sqrt{5}) = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}$.
$(ii)$ Using the identity $(a+b)(a-b) = a^2 - b^2$: $(5+\sqrt{5})(5-\sqrt{5}) = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20$.
$(iii)$ Using the identity $(a+b)^2 = a^2 + 2ab + b^2$: $(\sqrt{3}+\sqrt{7})^2 = (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{7}) + (\sqrt{7})^2 = 3 + 2\sqrt{21} + 7 = 10 + 2\sqrt{21}$.
$(iv)$ Using the identity $(a-b)(a+b) = a^2 - b^2$: $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7}) = (\sqrt{11})^2 - (\sqrt{7})^2 = 11 - 7 = 4$.

(A) To rationalise the denominator of $\frac{1}{\sqrt{2}}$,we multiply both the numerator and the denominator by $\sqrt{2}$.
$\frac{1}{\sqrt{2}} = \frac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
Since $\sqrt{2} \times \sqrt{2} = 2$,we get:
$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Thus,the rationalised form is $\frac{\sqrt{2}}{2}$.

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

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

(A) To rationalise the denominator of $\frac{1}{7+3 \sqrt{2}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(7-3 \sqrt{2})$.
$\frac{1}{7+3 \sqrt{2}} = \frac{1}{7+3 \sqrt{2}} \times \frac{7-3 \sqrt{2}}{7-3 \sqrt{2}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{7-3 \sqrt{2}}{(7)^2 - (3 \sqrt{2})^2}$
$= \frac{7-3 \sqrt{2}}{49 - (9 \times 2)}$
$= \frac{7-3 \sqrt{2}}{49 - 18}$
$= \frac{7-3 \sqrt{2}}{31}$

(N/A) $(i)$ $2-\sqrt{5}$: Since it is the difference between a rational number and an irrational number,it is an irrational number.
$(ii)$ $(3+\sqrt{23})-\sqrt{23} = 3+\sqrt{23}-\sqrt{23} = 3$. Since $3$ can be expressed as $\frac{3}{1}$,it is a rational number.
$(iii)$ $\frac{2 \sqrt{7}}{7 \sqrt{7}} = \frac{2}{7}$. Since this is in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$,it is a rational number.
$(iv)$ $\frac{1}{\sqrt{2}}$: The quotient of a rational number and an irrational number is always an irrational number. Therefore,it is an irrational number.
$(v)$ $2 \pi$: The product of a non-zero rational number and an irrational number is always an irrational number. Therefore,$2 \pi$ is an irrational number.

(C) To add the given expressions,we treat $\sqrt{3}$ as a common variable or term.
Given expression: $2 \sqrt{3} + \sqrt{3}$.
This can be written as $(2 + 1) \sqrt{3}$.
Calculating the sum inside the parentheses: $3 \sqrt{3}$.
Therefore,the correct option is $(c)$.

(D) Using the property of radicals $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$,we have:
$\sqrt{10} \times \sqrt{15} = \sqrt{10 \times 15}$
$= \sqrt{150}$
$= \sqrt{25 \times 6}$
$= \sqrt{25} \times \sqrt{6}$
$= 5 \sqrt{6}$
Therefore,the correct option is $D$.

(A) To rationalize the denominator of $\frac{1}{7-\sqrt{2}}$,we multiply both the numerator and the denominator by the conjugate of the denominator,which is $(7+\sqrt{2})$.
$\frac{1}{7-\sqrt{2}} = \frac{1}{7-\sqrt{2}} \times \frac{7+\sqrt{2}}{7+\sqrt{2}}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:
$= \frac{7+\sqrt{2}}{(7)^2 - (\sqrt{2})^2}$
$= \frac{7+\sqrt{2}}{49-2}$
$= \frac{7+\sqrt{2}}{47}$
Hence,option $A$ is the correct answer.

(B) Given expression: $\frac{1}{\sqrt{9}-\sqrt{8}}$
Since $\sqrt{9} = 3$ and $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$,the expression becomes:
$\frac{1}{3-2\sqrt{2}}$
To rationalize the denominator,multiply the numerator and denominator by the conjugate $(3+2\sqrt{2})$:
$\frac{1}{3-2\sqrt{2}} \times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} = \frac{3+2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2}$
Simplify the denominator:
$(3)^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1$
Thus,the expression simplifies to:
$\frac{3+2\sqrt{2}}{1} = 3+2\sqrt{2}$
Hence,option $(B)$ is the correct answer.

(C) To rationalize the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}},$ we multiply the numerator and the denominator by the conjugate of the denominator,which is $(3 \sqrt{3}+2 \sqrt{2}).$
$\frac{7}{3 \sqrt{3}-2 \sqrt{2}} \times \frac{3 \sqrt{3}+2 \sqrt{2}}{3 \sqrt{3}+2 \sqrt{2}} = \frac{7(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3})^2 - (2 \sqrt{2})^2}$
$= \frac{7(3 \sqrt{3}+2 \sqrt{2})}{(9 \times 3) - (4 \times 2)} = \frac{7(3 \sqrt{3}+2 \sqrt{2})}{27 - 8}$
$= \frac{7(3 \sqrt{3}+2 \sqrt{2})}{19}$
Thus,the denominator is $19.$
Therefore,the correct option is $(C)$.

(D) Given expression: $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$
Simplify the square roots in the numerator and denominator:
$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$
$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
Substitute these values back into the expression:
$\frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}$
Factor out the common terms:
$\frac{4(\sqrt{2} + \sqrt{3})}{2(\sqrt{2} + \sqrt{3})}$
Cancel the common term $(\sqrt{2} + \sqrt{3})$:
$\frac{4}{2} = 2$
Hence,the correct option is $(d)$.

(A) To simplify the expression $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$,we rationalize the denominator:
$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{\frac{(\sqrt{2}-1) \times (\sqrt{2}-1)}{(\sqrt{2}+1) \times (\sqrt{2}-1)}}$
$= \sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2 - 1^2}}$
$= \sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}$
$= \sqrt{(\sqrt{2}-1)^2}$
$= \sqrt{2}-1$
Given $\sqrt{2} = 1.4142$,we have:
$1.4142 - 1 = 0.4142$
Hence,the correct option is $(A)$.

(A) Yes,$x+y$ is necessarily an irrational number.
Proof by contradiction:
Suppose $x+y = r$,where $r$ is a rational number.
Since $x$ is a rational number,we can write $y = r - x$.
Since the difference of two rational numbers ($r$ and $x$) is always a rational number,$y$ must be a rational number.
However,this contradicts the given information that $y$ is an irrational number.
Therefore,our assumption is wrong,and $x+y$ must be an irrational number.
Example:
Let $x = 5$ (rational) and $y = \sqrt{2}$ (irrational).
Then,$x+y = 5 + \sqrt{2} = 6.4142...$,which is a non-terminating and non-repeating decimal.
Hence,$x+y$ is an irrational number.

(NO) Let $x = 0$ (a rational number) and $y = \sqrt{3}$ (an irrational number).
Then,the product $xy = 0 \times \sqrt{3} = 0$.
Since $0$ can be expressed as $\frac{0}{1}$,it is a rational number.
Therefore,$xy$ is not necessarily an irrational number.

(A) To simplify the expression $(3 \sqrt{5}-5 \sqrt{2})(4 \sqrt{5}+3 \sqrt{2})$,we use the distributive property ($FOIL$ method):
$= (3 \sqrt{5} \times 4 \sqrt{5}) + (3 \sqrt{5} \times 3 \sqrt{2}) - (5 \sqrt{2} \times 4 \sqrt{5}) - (5 \sqrt{2} \times 3 \sqrt{2})$
$= (12 \times 5) + (9 \sqrt{10}) - (20 \sqrt{10}) - (15 \times 2)$
$= 60 + 9 \sqrt{10} - 20 \sqrt{10} - 30$
$= (60 - 30) + (9 \sqrt{10} - 20 \sqrt{10})$
$= 30 - 11 \sqrt{10}$

(B) To solve $\frac{6}{3 \sqrt{2}-2 \sqrt{3}}$,we rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(3 \sqrt{2}+2 \sqrt{3})$.
$\frac{6}{3 \sqrt{2}-2 \sqrt{3}} = \frac{6(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2}-2 \sqrt{3})(3 \sqrt{2}+2 \sqrt{3})}$
Using the identity $(x-y)(x+y) = x^2 - y^2$,the denominator becomes $(3 \sqrt{2})^2 - (2 \sqrt{3})^2 = (9 \times 2) - (4 \times 3) = 18 - 12 = 6$.
So,the expression simplifies to $\frac{6(3 \sqrt{2}+2 \sqrt{3})}{6} = 3 \sqrt{2} + 2 \sqrt{3}$.
Comparing this with the given equation $3 \sqrt{2} - a \sqrt{3} = 3 \sqrt{2} + 2 \sqrt{3}$,we get $-a \sqrt{3} = 2 \sqrt{3}$.
Dividing both sides by $-\sqrt{3}$,we find $a = -2$.

(N/A) $(i)$ $(3+\sqrt{3})(2+\sqrt{2}) = 3(2+\sqrt{2}) + \sqrt{3}(2+\sqrt{2})$
$= 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$
$(ii)$ Using the identity $(a+b)(a-b) = a^2 - b^2$:
$(3+\sqrt{3})(3-\sqrt{3}) = (3)^2 - (\sqrt{3})^2 = 9 - 3 = 6$
$(iii)$ Using the identity $(a+b)^2 = a^2 + b^2 + 2ab$:
$(\sqrt{5}+\sqrt{2})^2 = (\sqrt{5})^2 + (\sqrt{2})^2 + 2(\sqrt{5})(\sqrt{2}) = 5 + 2 + 2\sqrt{10} = 7 + 2\sqrt{10}$
$(iv)$ Using the identity $(a-b)(a+b) = a^2 - b^2$:
$(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2}) = (\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3$

(N/A) $(i)$ $\frac{1}{\sqrt{7}} = \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7}$
$(ii)$ $\frac{1}{\sqrt{7}-\sqrt{6}} = \frac{1 \times (\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})} = \frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7}+\sqrt{6}}{7-6} = \sqrt{7}+\sqrt{6}$
$(iii)$ $\frac{1}{\sqrt{5}+\sqrt{2}} = \frac{1 \times (\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})} = \frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{\sqrt{5}-\sqrt{2}}{5-2} = \frac{\sqrt{5}-\sqrt{2}}{3}$
$(iv)$ $\frac{1}{\sqrt{7}-2} = \frac{1 \times (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} = \frac{\sqrt{7}+2}{(\sqrt{7})^2 - (2)^2} = \frac{\sqrt{7}+2}{7-4} = \frac{\sqrt{7}+2}{3}$

(A) To find three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$,we can multiply the numerator and denominator of both fractions by $(3+1) = 4$ or any larger number like $10$.
Multiplying by $10$:
$\frac{5}{7} = \frac{5 \times 10}{7 \times 10} = \frac{50}{70}$
$\frac{6}{7} = \frac{6 \times 10}{7 \times 10} = \frac{60}{70}$
Now,we can choose any three rational numbers between $\frac{50}{70}$ and $\frac{60}{70}$,such as $\frac{51}{70}, \frac{52}{70}, \text{ and } \frac{53}{70}$.
Thus,three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$ are $\frac{51}{70}, \frac{52}{70}, \text{ and } \frac{53}{70}$.

(C) To simplify the expression $\sqrt{45}-3 \sqrt{20}+4 \sqrt{5}$,we first factor the numbers inside the square roots:
$\sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5}$
$\sqrt{20} = \sqrt{4 \times 5} = 2 \sqrt{5}$
Now,substitute these values back into the expression:
$3 \sqrt{5} - 3(2 \sqrt{5}) + 4 \sqrt{5}$
$= 3 \sqrt{5} - 6 \sqrt{5} + 4 \sqrt{5}$
$= (3 - 6 + 4) \sqrt{5}$
$= 1 \sqrt{5} = \sqrt{5}$

(D) First,simplify the square roots:
$\sqrt{24} = \sqrt{4 \times 6} = 2 \sqrt{6}$
$\sqrt{54} = \sqrt{9 \times 6} = 3 \sqrt{6}$
Substitute these into the expression:
$\frac{2 \sqrt{6}}{8} + \frac{3 \sqrt{6}}{9} = \frac{\sqrt{6}}{4} + \frac{\sqrt{6}}{3}$
Find a common denominator,which is $12$:
$\frac{3 \sqrt{6}}{12} + \frac{4 \sqrt{6}}{12} = \frac{3 \sqrt{6} + 4 \sqrt{6}}{12} = \frac{7 \sqrt{6}}{12}$

(A) To simplify the expression $4 \sqrt{12} \times 7 \sqrt{6}$,we first factorize the numbers inside the square roots:
$4 \sqrt{12} = 4 \sqrt{4 \times 3} = 4 \times 2 \sqrt{3} = 8 \sqrt{3}$
$7 \sqrt{6} = 7 \sqrt{2 \times 3}$
Now,multiply the two expressions:
$8 \sqrt{3} \times 7 \sqrt{2 \times 3} = (8 \times 7) \times (\sqrt{3} \times \sqrt{3}) \times \sqrt{2}$
$= 56 \times 3 \times \sqrt{2}$
$= 168 \sqrt{2}$

(B) To simplify the expression $4 \sqrt{28} \div 3 \sqrt{7}$,we first express the division as a fraction:
$\frac{4 \sqrt{28}}{3 \sqrt{7}}$
Next,we simplify the square root of $28$ by factoring it as $28 = 4 \times 7$:
$\sqrt{28} = \sqrt{4 \times 7} = 2 \sqrt{7}$
Now,substitute this back into the expression:
$\frac{4 \times (2 \sqrt{7})}{3 \sqrt{7}}$
Cancel the common term $\sqrt{7}$ from the numerator and the denominator:
$\frac{4 \times 2}{3} = \frac{8}{3}$

(B) Given expression: $3 \sqrt{3}+2 \sqrt{27}+\frac{7}{\sqrt{3}}$
First,simplify $\sqrt{27}$:
$\sqrt{27} = \sqrt{9 \times 3} = 3 \sqrt{3}$
Substitute this into the expression:
$=3 \sqrt{3}+2(3 \sqrt{3})+\frac{7}{\sqrt{3}}$
$=3 \sqrt{3}+6 \sqrt{3}+\frac{7}{\sqrt{3}}$
Rationalize the denominator of the third term:
$\frac{7}{\sqrt{3}} = \frac{7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{7 \sqrt{3}}{3}$
Now add the terms:
$=3 \sqrt{3}+6 \sqrt{3}+\frac{7 \sqrt{3}}{3}$
$=9 \sqrt{3}+\frac{7 \sqrt{3}}{3}$
$= \sqrt{3} \left(9+\frac{7}{3}\right)$
$= \sqrt{3} \left(\frac{27+7}{3}\right)$
$= \frac{34}{3} \sqrt{3}$

(D) To simplify the expression $(\sqrt{3}-\sqrt{2})^{2}$,we use the algebraic identity $(a-b)^{2} = a^{2} + b^{2} - 2ab$.
Here,$a = \sqrt{3}$ and $b = \sqrt{2}$.
Substituting these values into the identity:
$(\sqrt{3}-\sqrt{2})^{2} = (\sqrt{3})^{2} + (\sqrt{2})^{2} - 2(\sqrt{3})(\sqrt{2})$
Since $(\sqrt{x})^{2} = x$ and $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$:
$= 3 + 2 - 2\sqrt{3 \times 2}$
$= 5 - 2\sqrt{6}$

(A) To simplify the expression $\sqrt[4]{81}-8 \sqrt[3]{216}+15 \sqrt[5]{32}+\sqrt{225}$,we evaluate each radical term individually:
$1$. $\sqrt[4]{81} = \sqrt[4]{3^{4}} = 3$
$2$. $\sqrt[3]{216} = \sqrt[3]{6^{3}} = 6$,so $8 \sqrt[3]{216} = 8 \times 6 = 48$
$3$. $\sqrt[5]{32} = \sqrt[5]{2^{5}} = 2$,so $15 \sqrt[5]{32} = 15 \times 2 = 30$
$4$. $\sqrt{225} = \sqrt{15^{2}} = 15$
Substituting these values back into the original expression:
$3 - 48 + 30 + 15$
$= (3 + 30 + 15) - 48$
$= 48 - 48 = 0$

(B) Given expression: $\frac{3}{\sqrt{8}} + \frac{1}{\sqrt{2}}$
First,simplify $\sqrt{8}$ as $\sqrt{4 \times 2} = 2\sqrt{2}$.
Substitute this into the expression: $\frac{3}{2\sqrt{2}} + \frac{1}{\sqrt{2}}$.
To add these fractions,find a common denominator,which is $2\sqrt{2}$.
$\frac{3}{2\sqrt{2}} + \frac{1 \times 2}{\sqrt{2} \times 2} = \frac{3}{2\sqrt{2}} + \frac{2}{2\sqrt{2}}$.
$= \frac{3 + 2}{2\sqrt{2}} = \frac{5}{2\sqrt{2}}$.
Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:
$= \frac{5 \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}} = \frac{5\sqrt{2}}{2 \times 2} = \frac{5\sqrt{2}}{4}$.

(C) To simplify the expression $\frac{2 \sqrt{3}}{3} - \frac{\sqrt{3}}{6}$,we first find a common denominator for the fractions,which is $6$.
$\frac{2 \sqrt{3}}{3} - \frac{\sqrt{3}}{6} = \frac{2 \sqrt{3} \times 2}{3 \times 2} - \frac{\sqrt{3}}{6}$
$= \frac{4 \sqrt{3}}{6} - \frac{\sqrt{3}}{6}$
$= \frac{4 \sqrt{3} - \sqrt{3}}{6}$
$= \frac{3 \sqrt{3}}{6}$
$= \frac{\sqrt{3}}{2}$

(D) To rationalise the denominator of $\frac{2}{3 \sqrt{3}}$,we multiply both the numerator and the denominator by $\sqrt{3}$:
$\frac{2}{3 \sqrt{3}} = \frac{2}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2 \sqrt{3}}{3 \times 3}$
$= \frac{2 \sqrt{3}}{9}$

(A) To rationalise the denominator of $\frac{\sqrt{40}}{\sqrt{3}}$,we multiply both the numerator and the denominator by $\sqrt{3}$.
$\frac{\sqrt{40}}{\sqrt{3}} = \frac{\sqrt{40} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$
Since $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$,we have:
$= \frac{2\sqrt{10} \times \sqrt{3}}{3}$
$= \frac{2\sqrt{30}}{3}$

(B) To rationalise the denominator of $\frac{3+\sqrt{2}}{4 \sqrt{2}}$,we multiply both the numerator and the denominator by $\sqrt{2}$:
$\frac{3+\sqrt{2}}{4 \sqrt{2}} = \frac{3+\sqrt{2}}{4 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$= \frac{\sqrt{2}(3+\sqrt{2})}{4 \times (\sqrt{2} \times \sqrt{2})}$
$= \frac{3 \sqrt{2} + 2}{4 \times 2}$
$= \frac{3 \sqrt{2} + 2}{8}$

(C) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{41}+5)$.
$\frac{16}{\sqrt{41}-5} = \frac{16}{\sqrt{41}-5} \times \frac{\sqrt{41}+5}{\sqrt{41}+5}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:
$= \frac{16(\sqrt{41}+5)}{(\sqrt{41})^2 - (5)^2}$
$= \frac{16(\sqrt{41}+5)}{41 - 25}$
$= \frac{16(\sqrt{41}+5)}{16}$
$= \sqrt{41}+5$

(D) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(2+\sqrt{3})$.
$\frac{2+\sqrt{3}}{2-\sqrt{3}} = \frac{2+\sqrt{3}}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator and $(a+b)^2 = a^2 + b^2 + 2ab$ in the numerator:
$= \frac{(2+\sqrt{3})^2}{2^2 - (\sqrt{3})^2}$
$= \frac{2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})}{4 - 3}$
$= \frac{4 + 3 + 4\sqrt{3}}{1}$
$= 7 + 4\sqrt{3}$

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{2}-\sqrt{3})$:
$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2}$
$= \frac{\sqrt{12} - \sqrt{18}}{2 - 3}$
$= \frac{\sqrt{12} - \sqrt{18}}{-1}$
$= \sqrt{18} - \sqrt{12}$
$= \sqrt{9 \times 2} - \sqrt{4 \times 3}$
$= 3\sqrt{2} - 2\sqrt{3}$

(B) To rationalise the denominator,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}+\sqrt{2})$.
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator and $(a+b)^2 = a^2 + b^2 + 2ab$ in the numerator:
$= \frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2}$
$= \frac{3 + 2 + 2\sqrt{3} \times \sqrt{2}}{3 - 2}$
$= \frac{5 + 2\sqrt{6}}{1}$
$= 5 + 2\sqrt{6}$

(C) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{5}+\sqrt{3})$.
$\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
$= \frac{3(\sqrt{5})^2 + 3\sqrt{15} + \sqrt{15} + (\sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2}$
$= \frac{3(5) + 4\sqrt{15} + 3}{5 - 3}$
$= \frac{15 + 4\sqrt{15} + 3}{2}$
$= \frac{18 + 4\sqrt{15}}{2}$
$= \frac{2(9 + 2\sqrt{15})}{2}$
$= 9 + 2\sqrt{15}$

(D) Given expression: $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$
Simplify the denominator: $\sqrt{48} = \sqrt{16 \times 3} = 4 \sqrt{3}$ and $\sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2}$.
So,the expression becomes: $\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}$.
To rationalise the denominator,multiply the numerator and denominator by the conjugate $(4 \sqrt{3}-3 \sqrt{2})$:
$= \frac{(4 \sqrt{3}+5 \sqrt{2})(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3}+3 \sqrt{2})(4 \sqrt{3}-3 \sqrt{2})}$
$= \frac{(4 \sqrt{3})^2 - (4 \sqrt{3})(3 \sqrt{2}) + (5 \sqrt{2})(4 \sqrt{3}) - (5 \sqrt{2})(3 \sqrt{2})}{(4 \sqrt{3})^2 - (3 \sqrt{2})^2}$
$= \frac{48 - 12 \sqrt{6} + 20 \sqrt{6} - 30}{48 - 18}$
$= \frac{18 + 8 \sqrt{6}}{30}$
Divide numerator and denominator by $2$:
$= \frac{9 + 4 \sqrt{6}}{15}$

(A) To solve the expression,we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator,which is $(7-4 \sqrt{3})$.
$L.H.S. = \frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}$
$= \frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{49 - 48}$
$= \frac{11 - 6 \sqrt{3}}{1} = 11 - 6 \sqrt{3}$
Given that $11 - 6 \sqrt{3} = a - 6 \sqrt{3}$,by comparing both sides,we get $a = 11$.

(NONE) To solve the expression,we rationalize the denominator:
$L.H.S. = \frac{3-\sqrt{5}}{3+2 \sqrt{5}} \times \frac{3-2 \sqrt{5}}{3-2 \sqrt{5}}$
$= \frac{9 - 6\sqrt{5} - 3\sqrt{5} + 2(5)}{3^2 - (2\sqrt{5})^2}$
$= \frac{9 - 9\sqrt{5} + 10}{9 - 20} = \frac{19 - 9\sqrt{5}}{-11}$
$= -\frac{19}{11} + \frac{9}{11}\sqrt{5}$
Comparing this with $a\sqrt{5} - \frac{19}{11}$,we get:
$a\sqrt{5} - \frac{19}{11} = \frac{9}{11}\sqrt{5} - \frac{19}{11}$
Therefore,$a = \frac{9}{11}$.

(C) Given expression: $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} = 2 - b\sqrt{6}$
Rationalize the denominator by multiplying the numerator and denominator by $(3\sqrt{2} + 2\sqrt{3})$:
$= \frac{(\sqrt{2}+\sqrt{3})(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})}$
$= \frac{3(\sqrt{2})^2 + 2\sqrt{6} + 3\sqrt{6} + 2(\sqrt{3})^2}{(3\sqrt{2})^2 - (2\sqrt{3})^2}$
$= \frac{3(2) + 5\sqrt{6} + 2(3)}{18 - 12}$
$= \frac{6 + 5\sqrt{6} + 6}{6} = \frac{12 + 5\sqrt{6}}{6} = 2 + \frac{5}{6}\sqrt{6}$
Comparing $2 + \frac{5}{6}\sqrt{6}$ with $2 - b\sqrt{6}$:
$-b = \frac{5}{6} \Rightarrow b = -\frac{5}{6}$

(D) Given equation: $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$
Rationalizing the denominators:
$\frac{(7+\sqrt{5})^2}{(7-\sqrt{5})(7+\sqrt{5})} - \frac{(7-\sqrt{5})^2}{(7+\sqrt{5})(7-\sqrt{5})} = a+\frac{7}{11} \sqrt{5} b$
Using $(a+b)^2 = a^2+b^2+2ab$ and $(a-b)^2 = a^2+b^2-2ab$:
$\frac{49+5+14\sqrt{5}}{49-5} - \frac{49+5-14\sqrt{5}}{49-5} = a+\frac{7}{11} \sqrt{5} b$
$\frac{54+14\sqrt{5}}{44} - \frac{54-14\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$
$\frac{54+14\sqrt{5}-54+14\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$
$\frac{28\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$
Simplifying the fraction $\frac{28}{44}$ by dividing by $4$:
$\frac{7\sqrt{5}}{11} = a+\frac{7}{11} \sqrt{5} b$
Comparing both sides,we get $a=0$ and $b=1$.

(A) Given that $a = 2 + \sqrt{3}$.
First,we find the value of $\frac{1}{a}$ by rationalizing the denominator:
$\frac{1}{a} = \frac{1}{2 + \sqrt{3}}$
Multiply the numerator and denominator by the conjugate $(2 - \sqrt{3})$:
$\frac{1}{a} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$
Using the identity $(x + y)(x - y) = x^2 - y^2$ in the denominator:
$\frac{1}{a} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2}$
$\frac{1}{a} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$
Now,calculate $a - \frac{1}{a}$:
$a - \frac{1}{a} = (2 + \sqrt{3}) - (2 - \sqrt{3})$
$a - \frac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3}$
$a - \frac{1}{a} = 2\sqrt{3}$

(B) To rationalise the denominator,multiply the numerator and the denominator by $\sqrt{3}$:
$\frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{4\sqrt{3}}{3}$
Substitute the value $\sqrt{3} = 1.732$:
$= \frac{4 \times 1.732}{3}$
$= \frac{6.928}{3}$
$= 2.309$

(C) To rationalise the denominator of $\frac{6}{\sqrt{6}}$,we multiply the numerator and the denominator by $\sqrt{6}$:
$\frac{6}{\sqrt{6}} = \frac{6 \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} = \frac{6\sqrt{6}}{6} = \sqrt{6}$
We know that $\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$.
Given $\sqrt{2} = 1.414$ and $\sqrt{3} = 1.732$,we have:
$\sqrt{6} = 1.414 \times 1.732 = 2.449048$
Rounding to three decimal places,we get $2.449$.

(D) Given expression: $\frac{\sqrt{10}-\sqrt{5}}{2}$
Step $1$: Simplify the numerator by factoring out $\sqrt{5}$:
$\frac{\sqrt{5} \times \sqrt{2} - \sqrt{5}}{2} = \frac{\sqrt{5}(\sqrt{2}-1)}{2}$
Step $2$: Substitute the given values $\sqrt{5} = 2.236$ and $\sqrt{2} = 1.414$:
$= \frac{2.236(1.414 - 1)}{2}$
Step $3$: Perform the subtraction inside the parenthesis:
$= \frac{2.236(0.414)}{2}$
Step $4$: Divide $2.236$ by $2$:
$= 1.118 \times 0.414$
Step $5$: Multiply the values:
$= 0.462852$
Rounding to three decimal places,we get $0.463$.

(A) To rationalise the denominator of $\frac{\sqrt{2}}{2+\sqrt{2}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(2-\sqrt{2})$.
$\frac{\sqrt{2}}{2+\sqrt{2}} = \frac{\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{\sqrt{2}(2-\sqrt{2})}{(2)^2 - (\sqrt{2})^2}$
$= \frac{2\sqrt{2} - 2}{4 - 2}$
$= \frac{2\sqrt{2} - 2}{2}$
$= \frac{2(\sqrt{2} - 1)}{2}$
$= \sqrt{2} - 1$
Given $\sqrt{2} = 1.414$,we substitute the value:
$= 1.414 - 1 = 0.414$