Question

plz help me....... ​

Answer 1

QUESTION :–

Simplify that :–

$\: { \dfrac{1}{ \sqrt{x + 2 \sqrt{x - 1} } } + \dfrac{1}{ \sqrt{x - 2 \sqrt{x - 1} } } }$

ANSWER :–

Given equation – $\: \\ { \dfrac{1}{ \sqrt{x + 2 \sqrt{x - 1} } } + \dfrac{1}{ \sqrt{x - 2 \sqrt{x - 1} } } }$

• Let's put ${ \bold{ \sqrt{x - 1} = t}}$ –

$\\ \: = { \dfrac{1}{ \sqrt{x + 2t } } + \dfrac{1}{ \sqrt{x - 2t} } }$

$\\ \: = \dfrac{ \sqrt{x - 2t} + \sqrt{x + 2t} }{( \sqrt{x + 2t})( \sqrt{x - 2t}) }$

$\\ \: = \dfrac{ \sqrt{ (\sqrt{x - 2t} + \sqrt{x + 2t} )^{2} }}{ \sqrt{ {x}^{2} - {(2t)}^{2} }}$

$\\ \: = \dfrac{ \sqrt{ (\sqrt{x - 2t})^{2} + (\sqrt{x + 2t} )^{2} + 2 (\sqrt{x - 2t} )( \sqrt{x + 2t} ) }}{ \sqrt{ {x}^{2} - {4t}^{2} }}$

$\\ \: = \dfrac{ \sqrt{x + 2t + x - 2t + 2 \sqrt{ {(x)}^{2} - {(2t)}^{2} } } }{ \sqrt{ {x}^{2} - 4 {t}^{2} } }$

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x}^{2} - {4t}^{2} }) } }{ \sqrt{ {x}^{2} - 4 {t}^{2} } }$

• Now put the value of t –

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x}^{2} - {4( \sqrt{x - 1} )}^{2} }) } }{ \sqrt{ {x}^{2} - 4 {( \sqrt{x - 1} )}^{2} } }$

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x}^{2} - {4(x - 1) }}} }{ \sqrt{ {x}^{2} - 4(x - 1) } }$

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x}^{2} - 4x + 4) }} }{ \sqrt{ {x}^{2} - 4x + 4} }$

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x - 2)}^{2} }} }{ \sqrt{ {(x - 2)}^{2} } }$

$\\ \: = \dfrac{ \sqrt{2x \pm 2 {(x - 2) }} }{ \pm(x - 2) }$

• Now take (+) sign –

$\\ \: = \dfrac{ \sqrt{2x + 2 {(x - 2) }} }{ (x - 2) }$

$\\ \: = \dfrac{ \sqrt{{4x - 4 }} }{ (x - 2) }$

$\\ \: = \dfrac{ 2\sqrt{{x - 1 }} }{ (x - 2) }$

=> but denominator will be positive –

$\\ \: = \dfrac{ 2\sqrt{{x - 1 }} }{ (x - 2) }$ {x > 2} [option (c)]

• Now take (-) sign –

$\\ \: = \dfrac{ \sqrt{2x - 2 {(x - 2) }} }{ - (x - 2) }$

$\\ \: = \dfrac{ \sqrt{4} }{ (2 - x) }$

$\\ \: = \dfrac{2}{ (2 - x) }$

=> but –

$\\ \implies \: 0 < \sqrt{x - 1} < 1$

$\\ \implies 1 < x < 2$

▪︎ So , that –

$\\ \: = \dfrac{2}{ (2 - x) }$ {1 < x < 2} [option (b)]

Hence , option (b) & (c) are correct.

Answer 2

QUESTION :–

Simplify that :–

$\: { \dfrac{1}{ \sqrt{x + 2 \sqrt{x - 1} } } + \dfrac{1}{ \sqrt{x - 2 \sqrt{x - 1} } } }$

ANSWER :–

Given equation – $\: \\ { \dfrac{1}{ \sqrt{x + 2 \sqrt{x - 1} } } + \dfrac{1}{ \sqrt{x - 2 \sqrt{x - 1} } } }$

• Let's put ${ \bold{ \sqrt{x - 1} = t}}$ –

$\\ \: = { \dfrac{1}{ \sqrt{x + 2t } } + \dfrac{1}{ \sqrt{x - 2t} } }$

$\\ \: = \dfrac{ \sqrt{x - 2t} + \sqrt{x + 2t} }{( \sqrt{x + 2t})( \sqrt{x - 2t}) }$

$\\ \: = \dfrac{ \sqrt{ (\sqrt{x - 2t} + \sqrt{x + 2t} )^{2} }}{ \sqrt{ {x}^{2} - {(2t)}^{2} }}$

$\\ \: = \dfrac{ \sqrt{ (\sqrt{x - 2t})^{2} + (\sqrt{x + 2t} )^{2} + 2 (\sqrt{x - 2t} )( \sqrt{x + 2t} ) }}{ \sqrt{ {x}^{2} - {4t}^{2} }}$

$\\ \: = \dfrac{ \sqrt{x + 2t + x - 2t + 2 \sqrt{ {(x)}^{2} - {(2t)}^{2} } } }{ \sqrt{ {x}^{2} - 4 {t}^{2} } }$

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x}^{2} - {4t}^{2} }) } }{ \sqrt{ {x}^{2} - 4 {t}^{2} } }$

• Now put the value of t –

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x}^{2} - {4( \sqrt{x - 1} )}^{2} }) } }{ \sqrt{ {x}^{2} - 4 {( \sqrt{x - 1} )}^{2} } }$

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x}^{2} - {4(x - 1) }}} }{ \sqrt{ {x}^{2} - 4(x - 1) } }$

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x}^{2} - 4x + 4) }} }{ \sqrt{ {x}^{2} - 4x + 4} }$

$\\ \: = \dfrac{ \sqrt{2x + 2 \sqrt{ {(x - 2)}^{2} }} }{ \sqrt{ {(x - 2)}^{2} } }$

$\\ \: = \dfrac{ \sqrt{2x \pm 2 {(x - 2) }} }{ \pm(x - 2) }$

• Now take (+) sign –

$\\ \: = \dfrac{ \sqrt{2x + 2 {(x - 2) }} }{ (x - 2) }$

$\\ \: = \dfrac{ \sqrt{{4x - 4 }} }{ (x - 2) }$

$\\ \: = \dfrac{ 2\sqrt{{x - 1 }} }{ (x - 2) }$

=> but denominator will be positive –

$\\ \: = \dfrac{ 2\sqrt{{x - 1 }} }{ (x - 2) }$ {x > 2} [option (c)]

• Now take (-) sign –

$\\ \: = \dfrac{ \sqrt{2x - 2 {(x - 2) }} }{ - (x - 2) }$

$\\ \: = \dfrac{ \sqrt{4} }{ (2 - x) }$

$\\ \: = \dfrac{2}{ (2 - x) }$

=> but –

$\\ \implies \: 0 < \sqrt{x - 1} < 1$

$\\ \implies 1 < x < 2$

▪︎ So , that –

$\\ \: = \dfrac{2}{ (2 - x) }$ {1 < x < 2} [option (b)]

Hence , option (b) & (c) are correct.