Question

for mi hater?

free points

rationalize the denominator 1/√3+√2​

Answer 1

Hii buddy

ur answer is

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

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

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

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

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

But I am the mi fan.....

Answer 2

$\huge\bold\star\red{Answer}$

rationalising the denominator 1/√3+√2

=> 1/√3+√2

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=> 1/√3+√2 × √3-√2/√3-√2

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=> √3+√2/1

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=> √3+√2

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$\huge\bold\star\red{thanx}$

$\huge{ \red{ \boxed{ \green{hope\:it \: helps:)}}}}$