Question

if root 5 equals to 2.236 and root 6 equals to 2.449 find the value of 1 plus root 2/ root 5 plus root 3 plus 1- root2/root 5- root 3

Answer 1

Hope u like my process
=====================
$\frac{ 1 + \sqrt{2} }{ \sqrt{5} + \sqrt{3} } + \frac{1 - \sqrt{2} }{ \sqrt{5} - \sqrt{3} } \\ = \frac{(1 + \sqrt{2})( \sqrt{5} - \sqrt{3}) + (1 - \sqrt{2} )( \sqrt{5} + \sqrt{3} )}{( \sqrt{5} + \sqrt{3})( \sqrt{5} - \sqrt{3} ) } \\ = \frac{ \sqrt{5} + \sqrt{10} - \sqrt{3} - \sqrt{6} + \sqrt{5} - \sqrt{10} + \sqrt{3} - \sqrt{6} }{ {( \sqrt{5} )}^{2} - {( \sqrt{3}) }^{2} } \\ = \frac{2 \sqrt{5 }- 2 \sqrt{6} }{5 - 3} = \frac{2( \sqrt{5} - \sqrt{6}) }{2} \\ = \sqrt{5} - \sqrt{6} \\ = 2.236 - 2.449 = - 0.213$
Hope this is ur required answer
Proud to help you

Answer 2

Answer:

Value of given expression is -0.213

Step-by-step explanation:

Given: √5 = 2.236 and √6 = 2.446

Consider,

$\frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}+\frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}$

rationalize the denominator of each term

$=\frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}\times\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}\times\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$=\frac{(1+\sqrt{2})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}+\frac{(1-\sqrt{2})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$

$=\frac{\sqrt{5}-\sqrt{3}+\sqrt{10}-\sqrt{6}}{(\sqrt{5})^2-(\sqrt{3})^2}+\frac{\sqrt{5}+\sqrt{3}-\sqrt{10}-\sqrt{6}}{(\sqrt{5})^2-(\sqrt{3})^2}$

$=\frac{\sqrt{5}-sqrt{3}+\sqrt{10}-\sqrt{6}+\sqrt{5}+\sqrt{3}-\sqrt{10}-\sqrt{6}}{5-3}$

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

$=\sqrt{5}-\sqrt{6}$

Now putting given value,

$=2.236-2.449$

$=-0.213$

Therefore, Value of given expression is -0.213