Question

Rationalize √3 + √2 + √5​

Answer 1

Answer:

Step-by-step explanation:

To Rationalize :-

√3 + √2 + √5​

Solution :-

Multiplying and dividing with rationalizing factor '√3 + √2 - √5​' :-

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

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

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

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

$=\dfrac{3+2+2\sqrt{6}-5}{\sqrt{3}\sqrt{2}-\sqrt{5}}$

$=\dfrac{5-5 +2\sqrt{6}}{\sqrt{3}+\sqrt{2}-\sqrt{5}}$

$=\dfrac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}-\sqrt{5}}$

Multiplying and dividing with rationalizing  factor '√6' :-

$=\dfrac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}-\sqrt{5}}\times \dfrac{\sqrt{6}}{\sqrt{6}}$

$=\dfrac{2\times (\sqrt{6})^2}{\sqrt{6}(\sqrt{3}+\sqrt{2}-\sqrt{5})}$

$=\dfrac{2\times 6}{\sqrt{18}+\sqrt{12}-\sqrt{30}}$

$=\dfrac{12}{\sqrt{18}+\sqrt{12}-\sqrt{30}}$

Answer 2

$\huge \mathfrak \red{Answer :}$

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

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

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

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

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

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

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

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

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

$= \frac{2 \times 6}{ \sqrt{18} + \sqrt{12} - \sqrt{30} }$

$= \large \tt \green{\frac{12}{ \sqrt{18} + \sqrt{12} - \sqrt{30} } }$

For :- prashansa2008