Question

solve it and i will mark as brainliest​

Answer 1

Step-by-step explanation:

x 2 +1 =log{ x 2 +1 −x}

Differentiating with reset to be x we get

Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:2 \sqrt{x} + \sqrt{5x - 4 \sqrt{x} } = 1$

$\large\underline{\sf{To\:show - }}$

$\rm :\longmapsto\:x - 1 = 0$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:2 \sqrt{x} + \sqrt{5x - 4 \sqrt{x} } = 1$

can be rewritten as

$\rm :\longmapsto\: \sqrt{5x - 4 \sqrt{x} } = 1 - 2 \sqrt{x}$

On squaring both sides, we get

$\rm :\longmapsto\: (\sqrt{5x - 4 \sqrt{x} })^{2} = (1 - 2 \sqrt{x} )^{2}$

We know,

$\boxed{ \rm{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2yx}}$

Using this identity, we get .

$\rm :\longmapsto\:5x - 4 \sqrt{x} = ( {1)}^{2} + {(2 \sqrt{x}) }^{2} - 2 \times 1 \times 2 \sqrt{x}$

$\rm :\longmapsto\:5x - 4 \sqrt{x} = 1 + 4x - 4 \sqrt{x}$

$\rm :\longmapsto\:5x = 1 + 4x$

$\rm :\longmapsto\:5x - 4x = 1$

$\rm :\longmapsto\:x - 1 = 0$

Hence, proved

Additional Information :-

Let's solve one more problem of same type!!

Question :-

$\rm{ \: If \: \sqrt{x + 1} + \sqrt{x - 2} = 3}, \: show \: that \: x - 3 = 0$

Answer :-

Given that

$\rm :\longmapsto\: \rm{ \: \: \sqrt{x + 1} + \sqrt{x - 2} = 3}$

can be rewritten as

$\rm :\longmapsto\: \sqrt{x - 2} = 3 - \sqrt{x + 1}$

On squaring both sides, we get

$\rm :\longmapsto\: (\sqrt{x - 2} )^{2} = (3 - \sqrt{x + 1} )^{2}$

$\rm :\longmapsto\:x - 2 = 9 + (x + 1) - 6 \sqrt{x + 1}$

$\rm :\longmapsto\:x - 2 = 9 + x + 1 - 6 \sqrt{x + 1}$

$\rm :\longmapsto\:x - 2 = x + 10- 6 \sqrt{x + 1}$

$\rm :\longmapsto\:- 2 = 10- 6 \sqrt{x + 1}$

$\rm :\longmapsto\:- 2 - 10 = - 6 \sqrt{x + 1}$

$\rm :\longmapsto\:- 12 = - 6 \sqrt{x + 1}$

$\rm :\longmapsto\:2 = \sqrt{x + 1}$

On squaring both sides, we get

$\rm :\longmapsto\:4 = x + 1$

$\bf\implies \:x - 3 = 0$