Question

rationalise the denominator-
$\frac{ \sqrt{a + x} - \sqrt{a - x} }{ \sqrt{a + x} + \sqrt{a - x} }$
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Answer 1

Answer:

Please see the attachment

Answer 2

Answer:

${\sf{ {\dfrac{a - {\sqrt{a^2 - x^2}} }{x}}}}$

Step-by-step explanation:

Given : ${\sf{\ \ {\dfrac{ {\sqrt{a + x}} - {\sqrt{a - x}} }{ {\sqrt{a + x}} + {\sqrt{a - x}}}} }}$

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Rationalising the denominator.

$\implies{\sf{ {\dfrac{ {\sqrt{a + x}} - {\sqrt{a - x}} }{ {\sqrt{a + x}} + {\sqrt{a - x}}}} \times {\dfrac{ {\sqrt{a + x}} - {\sqrt{a - x}} }{ {\sqrt{a + x}} - {\sqrt{a - x}} }} }}$

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$\implies{\sf{ {\dfrac{ ( {\sqrt{a + x}} - {\sqrt{a - x}} )^2 }{ ( {\sqrt{a + x}} + {\sqrt{a - x}} )( {\sqrt{a + x}} - {\sqrt{a - x}} ) }}}}$

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${\boxed{\tt{\bigstar \ \ Identity \ : \ (a + b)(a - b) = a^2 - b^2}}}$

${\tt{Here, \ a = ( {\sqrt{a + x}} ), \ b = ( {\sqrt{a - x}} )}}$

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${\boxed{\tt{\bigstar \ \ Identity \ : \ (a - b)^2 = a^2 + b^2 - 2ab}}}$

${\tt{Here, \ a = ( {\sqrt{a + x}} ), \ b = ( {\sqrt{a - x}} )}}$

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$\implies{\sf{ {\dfrac{ ( {\sqrt{a + x}} )^2 + ( {\sqrt{a - x}} )^2 - 2 ( {\sqrt{a + x}} )( {\sqrt{a - x}} ) }{ ( {\sqrt{a + x}} )^2 - ( {\sqrt{a - x}} )^2 }}}}$

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$\implies{\sf{ {\dfrac{ a + x + a - x - 2 {\sqrt{(a + x)(a - x)}} }{ (a + x) - (a - x) }}}}$

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$\implies{\sf{ {\dfrac{ a + x + a - x - 2 {\sqrt{(a + x)(a - x)}} }{ a + x - a + x }}}}$

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$\implies{\sf{ {\dfrac{ a + a - 2 {\sqrt{(a + x)(a - x)}} }{ x + x }}}}$

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${\boxed{\tt{\bigstar \ \ Identity \ : \ (a + b)(a - b) = a^2 - b^2}}}$

${\tt{Here, \ a = a, \ b = x}}$

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$\implies{\sf{ {\dfrac{ a + a - 2 {\sqrt{(a)^2 - (x)^2}} }{ x + x }}}}$

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$\implies{\sf{ {\dfrac{ 2a - 2 {\sqrt{a^2 - x^2}} }{ 2x }}}}$

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We can write this as :

$\implies{\sf{ {\dfrac{ 2(a - {\sqrt{a^2 - x^2}}) }{ 2(x) }}}}$

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$\implies{\boxed{\boxed{\sf{ {\dfrac{a - {\sqrt{a^2 - x^2}} }{x}}}}}}$

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