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 $3$ to both sides of the equation:

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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 $3$ to both sides of the equation:$(\frac{p + q - x}{r} + 1) + (\frac{q + r - x}{p} + 1) + (\frac{r + p - x}{q} + 1) + \frac{3x}{p + q + r} = 3$$\frac{p + q + r - x}{r} + \frac{q + r + p - x}{p} + \frac{r + p + q - x}{q} + \frac{3x}{p + q + r} = 3$$(p + q + r - x) (\frac{1}{r} + \frac{1}{p} + \frac{1}{q}) = 3 - \frac{3x}{p + q + r}$$(p + q + r - x) (\frac{1}{p} + \frac{1}{q} + \frac{1}{r}) = 3 (\frac{p + q + r - x}{p + q + r})$$(p + q + r - x) [\frac{1}{p} + \frac{1}{q} + \frac{1}{r} - \frac{3}{p + q + r}] = 0$This implies $p + q + r - x = 0$ or the bracket term is zero.Therefore,$x = p + q + r$.

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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