The value of 3+1/(3)+1/(3+3)+1/(3-3) is Option 1: 3+3 Option 2: 3 Option 3: 1 Option 4: 0

Answer: 3. Solution: Add the last two fractions first, since those are the messy ones. 1/(3+3)+1/(3-3)=((3-3)+(3+3))/((3+3)(3-3)) Numerator=23 Using the identity (A+B)(A-B)=A^2-B^2 on the denominator: Denominator=(3)^2-3^2=3-9=-6 So their sum is (23)/(-6)=-(3)/3=-1/(3) Put this back into the original expression: 3+1/(3)-1/(3)=3

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Medium

The value of $3+\frac{1}{\sqrt3}+\frac{1}{3+\sqrt3}+\frac{1}{\sqrt3-3}$ is

$3+\sqrt3$

✅ Correct Option: 2

Add the last two fractions first, since those are the messy ones. $\dfrac{1}{3+\sqrt3}+\dfrac{1}{\sqrt3-3}=\dfrac{(\sqrt3-3)+(3+\sqrt3)}{(3+\sqrt3)(\sqrt3-3)}$ $\text{Numerator}=2\sqrt3$ Using the identity $(A+B)(A-B)=A^2-B^2$ on the denominator: $\text{Denominator}=(\sqrt3)^2-3^2=3-9=-6$ So their sum is $\dfrac{2\sqrt3}{-6}=-\dfrac{\sqrt3}{3}=-\dfrac{1}{\sqrt3}$ Put this back into the original expression: $3+\dfrac{1}{\sqrt3}-\dfrac{1}{\sqrt3}=3$

JIPMAT 2026

Arrange the following in descending order:

A. $\sqrt{401}-\sqrt{399}$

B. $\sqrt{301}-\sqrt{299}$

C. $\sqrt{101}-\sqrt{99}$

D. $\sqrt{201}-\sqrt{199}$

Choose the correct answer from the options given below:

JIPMAT 2025

$\sqrt{2+\sqrt{3}} \times \sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$ is equal to

JIPMAT 2021

If $2^{x} = 3^{y} = 6^{-z}$ , then $(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is equal to

JIPMAT 2023

$\sqrt{3+\sqrt{5}}=$

JIPMAT 2022

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2 \sqrt{15}}}+\frac{\sqrt{11+2 \sqrt{30}}}{\sqrt{6}-\sqrt{5}}$

JIPMAT 2024

Simplify : $\dfrac{\sqrt[4]{0.0625}+\sqrt[3]{0.008}+\sqrt{0.09}-1}{\sqrt[3]{62 \cdot 5 \times \sqrt[5]{32}}}$

The value of $3+\frac{1}{\sqrt3}+\frac{1}{3+\sqrt3}+\frac{1}{\sqrt3-3}$ is

Solution

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