Q.Solve:
\(\cfrac{a}{ax - 1} + \cfrac{b}{bx - 1} = a + b\),
\([x \ne \cfrac{1}{a}, \cfrac{1}{b}]\)

\(\cfrac{a}{ax - 1} + \cfrac{b}{bx - 1} = a + b\)
Or, \(\cfrac{a}{ax - 1} - b + \cfrac{b}{bx - 1} - a = 0\)
Or, \(\cfrac{a - abx + b}{ax - 1} + \cfrac{b - abx + a}{bx - 1} = 0\)
Or, \((a + b - abx)\Big[\cfrac{1}{ax - 1} + \cfrac{1}{bx - 1}\Big] = 0\)
Or, \((a + b - abx)\Big[\cfrac{bx - 1 + ax - 1}{(ax - 1)(bx - 1)}\Big] = 0\)
Or, \((a + b - abx)\Big[\cfrac{bx + ax - 2}{(ax - 1)(bx - 1)}\Big] = 0\)

Either, \((a + b - abx) = 0\)
Or, \(\cfrac{bx + ax - 2}{(ax - 1)(bx - 1)} = 0\)

When, \((a + b - abx) = 0\)
Then, \( -abx = -a - b\)
Or, \(x = \cfrac{a + b}{ab}\)
When, \(\cfrac{bx + ax - 2}{(ax - 1)(bx - 1)} = 0\)
Then, \( bx + ax - 2 = 0\)
Or, \(x = \cfrac{2}{a + b}\)

\(\therefore\) The roots of the quadratic equation are \(\cfrac{a + b}{ab}\) and \(\cfrac{2}{a + b}\).
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\([x\ne \cfrac{1}{a}, \cfrac{1}{b}]\) ">