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Mathematics

If x = 2maba+b,\dfrac{2mab}{a + b}, find the value of

x+max−ma+x+mbx−mb.\dfrac{x + ma}{x - ma} + \dfrac{x + mb}{x - mb}.

Ratio Proportion

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Answer

x=2maba+bx = \dfrac{2mab}{a + b}

Putting this value of x in

=x+max−ma+x+mbx−mb=2maba+b+ma2maba+b−ma+2maba+b+mb2maba+b−mb=2mab+ma2+maba+b2mab−ma2−maba+b+2mab+mb2+maba+b2mab−mb2−maba+b=2mab+ma2+mab2mab−ma2−mab+2mab+mb2+mab2mab−mb2−mab=ma(2b+a+b)ma(2b−a−b)+mb(2a+b+a)mb(2a−b−a)=3b+ab−a+3a+ba−b=3b+ab−a−3a+bb−a=3b+a−3a−bb−a=2b−2ab−a=2(b−a)(b−a)=2.\phantom{= }\dfrac{x + ma}{x - ma} + \dfrac{x + mb}{x - mb} \\[1.5em] = \dfrac{\dfrac{2mab}{a + b} + ma}{\dfrac{2mab}{a + b} - ma} + \dfrac{\dfrac{2mab}{a + b} + mb}{\dfrac{2mab}{a + b} - mb} \\[1em] = \dfrac{\dfrac{2mab + ma^2 + mab}{a + b}}{\dfrac{2mab - ma^2 - mab}{a + b}} + \dfrac{\dfrac{2mab + mb^2 + mab}{a + b}}{\dfrac{2mab - mb^2 - mab}{a + b}} \\[1em] = \dfrac{2mab + ma^2 + mab}{2mab - ma^2 - mab} + \dfrac{2mab + mb^2 + mab}{2mab - mb^2 - mab} \\[1em] = \dfrac{ma(2b + a + b)}{ma(2b - a - b)} + \dfrac{mb(2a + b + a)}{mb(2a - b - a)} \\[1em] = \dfrac{3b + a}{b - a} + \dfrac{3a + b}{a - b} \\[1em] = \dfrac{3b + a}{b - a} - \dfrac{3a + b}{b - a} \\[1em] = \dfrac{3b + a - 3a - b}{b - a} \\[1em] = \dfrac{2b - 2a}{b - a} \\[1em] = \dfrac{2(b - a)}{(b - a)} \\[1em] = 2.

Hence, the value of x+max−ma+x+mbx−mb=2.\dfrac{x + ma}{x - ma} + \dfrac{x + mb}{x - mb} = 2.

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