Question

If a=x/x+y and b=x/x-y, then ab/a+b is what

Answer 1

ANSWER:-

$given \: a = \frac{x}{x + y} \\ \: \: \: \: \: \: \: \: \: \: \: b = \frac{x}{x - y}$

REQUIRED TO FIND :

$\Huge \bf \frac {ab}{a+b}$

METHOD:

let us find both ab and a+b separately

$ab = \frac{x}{x + y} \times \frac{x}{x - y}$

$\Huge \bf {\underline{(x+y)(x-y)=x^2-y^2}}$

using the identity

$ab = \frac{ {x}^{2} }{ {x}^{2} - {y}^{2} }$

let us

now find a+b

$a + b = \frac{x}{x + y} + \frac{x}{x - y}$

$a + b = \frac{ {x}^{2} - yx + {x}^{2} + yx }{ {x}^{2} - {y}^{2} }$

$= \frac{2 {x}^{2} }{ {x}^{2} - {y}^{2} }$

now the value of ab/a+b is

$\frac{ab}{a + b} = \frac{ \frac{ {x}^{2} }{ {x}^{2} - {y}^{2} } }{ \frac{2 {x}^{2} }{ {x}^{2} - {y}^{2} } }$

$= \frac{ {x}^{2} }{ {x}^{2} - {y}^{2} } \times \frac{ {x}^{2} - {y}^{2} }{2 {x}^{2} }$

$= \frac{1}{2}$

$so \:the \:value \:of \: \frac {ab}{a+b} \: is$

$\Huge \bf {\underline{\frac{1}{2}}}$

IDENTITY USED:

(a+b)(a-b)=a^2-b^2

Answer 2

Answer:

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Step-by-step explanation:

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