QQuestion
Solve: \frac{1}{\text{a+b+x}} = \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{x}}, x ≠ 0, -(a + b)
✓Answer
Given: \frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x} with x ≠ 0, – (a + b)
or, \frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a}+\frac{1}{b}
or, \frac{x - a - b - x}{(a+b+x)x} = \frac{a + b}{ab}
or, – \frac{(a + b)}{(a+b+x)x} = \frac{a + b}{ab}
or, – \frac{1}{(a+b+x)x} = \frac{1}{ab}
or, – ab = (a+b+x)x
or, x2 + ax + bx + ab = 0
or, x (x + a) + b (x + a) = 0
or, (x + a)(x + b) = 0
x = -a or -b
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