Question

please answer this question​

Answer 1

Answer:

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Answer 2

$\huge\underline{\fcolorbox{yellow}{pink}{Answer}}$

$\frac{1}{p + q + x} = \frac{1}{p} + \frac{1}{q } + \frac{1}{x} \\ </p><p></p><p> = > \frac{1}{p + q+x } - \frac{1}{x} = \frac{1}{p} + \frac{1}{q} \\ </p><p></p><p>= > \frac{x - p - q - x}{px + qx + {x}^{2} } = \frac{q+ p}{pq} \\ </p><p></p><p> = > \frac{ - (p + q)}{ {x}^{2} + px + qx} = \frac{p + q}{pq} \\ </p><p></p><p>= > \frac{ - 1}{ {x}^{2} + px + qx} = \frac{1}{pq} \\ </p><p></p><p>= > - pq = {x}^{2} + px + qx \\ </p><p></p><p>= > {x}^{2} + px + qx + pq = 0 \\ </p><p></p><p> = > x(x + p) + q( x + p) = 0 \\ </p><p></p><p>= > (x + p)(x + q) = 0 \\ so \\ x + p = 0 \: \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: \: \: x + q = 0 \\ = > x = - p \: \: \: \: \: \: \: \: \: \: \: \: = > x = - q$

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