Question

ans it ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤​

Answer 1

Answer:

Solution☺:-

$</p><p></p><p>\begin{gathered}\frac{a}{ax + 1} + \frac{b}{bx - 1} = a + b \\ = > ( \frac{a}{ax - 1} - b) + ( \frac{b}{bx - a} ) = 0 \\ = > \frac{a - abx + b}{ax - 1} + \frac{b - abx + a}{bx - 1} = 0 \\ = > (a + b - abx)( \frac{1}{ax - 1} + \frac{1}{bx - 1} = 0 \\ = > (a + b - abx)( \frac{bx - 1 + ax - 1}{(ax - 1)(bx - 1)} = 0 \\ = > (a + b - abx)(ax + bx - 2) = 0 \\ either \\ a + b - abx = 0 \\ = > a+ b = abx \\ = > x = \frac{a + b}{ab} \\ \\ or \\ ax + bx - 2 = 0 \\ = > x(a + b) = 2 \\ = > x = \frac{2}{a +b }\end{gathered} </p><p>$

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Answer 2

Answer:

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