Монгол Бодлогын Сан

Заавар: $\dfrac{\frac{2}{x}-\frac{1}{y+z}}{\frac{2}{x}+\frac{1}{y+z}}=\dfrac{2(y+z)-x}{2(y+z)+x}$, $1+\dfrac{4(y^2+z^2)-x^2}{8yz}=\dfrac{4(y+z)^2-x^2}{8yz}=\dfrac{(2(y+z)-x)(2(y+z)+x)}{8yz}$.

Бодолт: $$\begin{align*} \text{Илэрх.}&=\dfrac{\frac{2}{x}-\frac{1}{y+z}}{\frac{2}{x}+\frac{1}{y+z}}\cdot\Big(1+\dfrac{4(y^2+z^2)-x^2}{8yz}\Big):\dfrac{x-2(y+z)}{xyz}\\ &=\dfrac{\bcancel{2(y+z)-x}^{(-1}}{\xcancel{2(y+z)+x}}\cdot\dfrac{(2(y+z)-x)\xcancel{(2(y+z)+x)}}{8\cancel{yz}}\cdot\dfrac{x\cancel{yz}}{\bcancel{x-2(y+z)}}\\ &=\dfrac{-(2(y+z)-x)x}{8}=\dfrac{x(x-2(y+z))}{8} \end{align*}$$