Question

simplify 2 root 3 by root 3-root 2 +3 root 2 by root 3 + root 2

Answer 1

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Answer 2

Answer:

$\frac{2\sqrt3}{\sqrt3-\sqrt2}+\frac{3\sqrt2}{\sqrt3+\sqrt2}=5\sqrt 6$

Step-by-step explanation:

Given : Expression $\frac{2\sqrt3}{\sqrt3-\sqrt2}+\frac{3\sqrt2}{\sqrt3+\sqrt2}$

To find : Simplify the expression ?

Solution :

Expression $\frac{2\sqrt3}{\sqrt3-\sqrt2}+\frac{3\sqrt2}{\sqrt3+\sqrt2}$

Take LCM,

$=\frac{2\sqrt3(\sqrt3+\sqrt2)+3\sqrt2(\sqrt3-\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}$

$=\frac{2\times 3+2\sqrt6+3\sqrt6-3\times 2}{\sqrt3^2-\sqrt2^2}$

$=\frac{6+2\sqrt6+3\sqrt6-6}{3-2}$

$=\frac{2\sqrt6+3\sqrt6}{1}$

$=5\sqrt6$

Therefore, $\frac{2\sqrt3}{\sqrt3-\sqrt2}+\frac{3\sqrt2}{\sqrt3+\sqrt2}=5\sqrt 6$