Question

2 root3- root5/2 root2+3 root3

Answer 1

Answer:

$\frac{2\sqrt{3}-5\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}$

=$\frac{(4\sqrt{6}-18-10\sqrt{10}+15\sqrt{15})}{-19}$

Explanation:

Given

$\frac{2\sqrt{3}-5\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}$

Rationalising the denominator, we get

= $\frac{(2\sqrt{3}-5\sqrt{5})(2\sqrt{2}-3\sqrt{3})}{(2\sqrt{2}+3\sqrt{3})(2\sqrt{2}-3\sqrt{3})}$

=$\frac{(4\sqrt{6}-18-10\sqrt{10}+15\sqrt{15})}{(2\sqrt{2})^{2}-(3\sqrt{3})^{2}}$

=$\frac{(4\sqrt{6}-18-10\sqrt{10}+15\sqrt{15})}{8-27}$

=$\frac{(4\sqrt{6}-18-10\sqrt{10}+15\sqrt{15})}{-19}$

Therefore,

$\frac{2\sqrt{3}-5\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}$

=$\frac{(4\sqrt{6}-18-10\sqrt{10}+15\sqrt{15})}{-19}$

••••

Answer 2

The value of $\dfrac{2\sqrt{3}-\sqrt{5} }{2\sqrt{2}+3\sqrt{3} }$  is $\dfrac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{-19}$.

Step-by-step explanation:

We need to simplify the given expression :

$\dfrac{2\sqrt{3}-\sqrt{5} }{2\sqrt{2}+3\sqrt{3} }$

Rationalizing both numerator and denominator, such that,

$\dfrac{2\sqrt{3}-\sqrt{5} }{2\sqrt{2}+3\sqrt{3} }\times \dfrac{2\sqrt{2}-3\sqrt{3}}{2\sqrt{2}-3\sqrt{3}}$

$=\dfrac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{(2\sqrt{2})^2-(3\sqrt{3})^2 }$

$=\dfrac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{-19}$

So, the value of $\dfrac{2\sqrt{3}-\sqrt{5} }{2\sqrt{2}+3\sqrt{3} }$  is $\dfrac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{-19}$.

Learn more,

Simplification

https://brainly.in/question/4380660