The solution of the equation frac{p + q - x}{ | vedclass

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(A) Given equation: $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$Add $1$ to each of the first three ter

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$x = p + q + r$

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(A) Given equation: $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$Add $1$ to each of the first three terms to adjust the numerators:$(\frac{p + q - x}{r} + 1) + (\frac{q + r - x}{p} + 1) + (\frac{r + p - x}{q} + 1) + \frac{3x}{p + q + r} - 3 = 0$$\frac{p + q + r - x}{r} + \frac{p + q + r - x}{p} + \frac{p + q + r - x}{q} + \frac{3x - 3(p + q + r)}{p + q + r} = 0$$(p + q + r - x) (\frac{1}{r} + \frac{1}{p} + \frac{1}{q}) - \frac{3(p + q + r - x)}{p + q + r} = 0$$(p + q + r - x) [\frac{1}{p} + \frac{1}{q} + \frac{1}{r} - \frac{3}{p + q + r}] = 0$This implies either $(p + q + r - x) = 0$ or the bracketed term is zero.Thus,$x = p + q + r$.

The solution of the equation $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$ is

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