Question

Prove the following​

Answer 1

Step-by-step explanation:

Given -

1/√2-1 + 2/√3+1 = √2 + √3

To Show -

LHS = RHS

Now,

Rationalising the denominator -

1/√2-1 × √2+1/√2+1

= √2+1/2-1

• = √2 + 1

And

2/√3+1 × √3-1/√3-1

= 2(√3-1)/3-1

= 2(√3-1)/2

• = √3 - 1

Now,

√2 + 1 + √3 - 1

= √2 + √3

LHS = RHS

Hence,

Proved..

Answer 2

QuEsTiOn :-

Prove that

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

$\huge{ \underline{ \underline{ \green{ \sf{ PrOoF:- }}}}}$

Taking LHS :-

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

Rationalising the denominator =>

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

Identity used :-

$(a - b)(a + b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ \sqrt{2} + 1}{ 2 - 1} + \frac{2 \sqrt{3} - 2 }{3 - 1} \\ \\ = \frac{ \sqrt{2} + 1 }{1} + \frac{2( \sqrt{3} - 1) }{2} \\ \\ = \sqrt{2} + 1 + \sqrt{3} - 1 \\ \\ = \sqrt{2} + \sqrt{3} = RHS \\ \\ {\green{\boxed{\large{\bold{LHS \: = \: RHS}}}}}$

HeNcE PrOvEd