The square root of left[frac{(3.25)^{3}}{3.25 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $a = 3.25$.The given expression is $E = \left[\frac{a^3}{a-1} - (a^2 + a + 1)\right]$.We know that $a^3 - 1 = (a - 1)(a^2 + a + 1)$,which impl

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) Let $a = 3.25$.The given expression is $E = \left[\frac{a^3}{a-1} - (a^2 + a + 1)\right]$.We know that $a^3 - 1 = (a - 1)(a^2 + a + 1)$,which implies $\frac{a^3 - 1}{a - 1} = a^2 + a + 1$.Substituting this into the expression $E$:$E = \frac{a^3}{a-1} - \frac{a^3 - 1}{a-1} = \frac{a^3 - (a^3 - 1)}{a-1} = \frac{1}{a-1}$.Now,substitute $a = 3.25$:$E = \frac{1}{3.25 - 1} = \frac{1}{2.25} = \frac{1}{9/4} = \frac{4}{9}$.The square root of the expression is $\sqrt{E} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

The square root of $\left[\frac{(3.25)^{3}}{3.25-1}-\left(3.25+(3.25)^{2}+1\right)\right]$ is equal to

Similar Questions

Two baskets together have $640$ oranges. If $\frac{1}{5}$ of the oranges in the first basket are taken to the second basket,the number of oranges in both baskets becomes equal. The number of oranges in the first basket is:

$6894.986 + 5025.005 + 600.020 = ?$

The difference between the squares of two consecutive numbers is $35.$ The numbers are

The product of three consecutive even numbers is $4032$. The product of the first and the third number is $252$. What is five times the second number?

Find the greatest fraction among the fractions $\frac{17}{18}, \frac{21}{22}, \frac{26}{27}$ and $\frac{36}{37}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module