Алгебра, вопрос задал spookyghost , 2 года назад

помогите решить систему уравнений

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Ответил Аноним

$\displaystyle \begin{cases}\displaystyle \frac{1}{x+y} - \frac{6}{x-y} = -2 \\\\ \displaystyle\frac{1}{x+y} + \frac{3}{x-y} = \frac{1}{4}\displaystyle \end{cases} \\\\ ---------\\\\\displaystyle \frac{1}{x+y} = t \; \; , \; \; \frac{1}{x-y} = u \\\\\displaystyle \begin{cases}\displaystyle t-6u=-2 \\\\ \displaystyle\displaystyle t+3u=\frac{1}{4} \; \; \mid \; \; \cdot (-1)\displaystyle \end{cases} \\\\\displaystyle \begin{cases}\displaystyle t-6u=-2 \\\\ \displaystyle\displaystyle -t-3u=-\frac{1}{4}\displaystyle \end{cases} \\\\\displaystyle -9u=-\frac{9}{4} \;\; \mid \; \; \div (-9) \\\\\displaystyle u=\frac{1}{4} \\\\--------\\\\\displaystyle t + 3u = \frac{1}{4}\\\\\displaystyle t+3\cdot \frac{1}{4} = \frac{1}{4} \\\\\displaystyle t = -\frac{1}{2} \\\\--------\\\\ \displaystyle \begin{cases}\displaystyle -\frac{1}{2} - 6\cdot \frac{1}{4}=-2 \\\\ \displaystyle-\frac{1}{2}+3\cdot \frac{1}{4}=\frac{1}{4}\displaystyle \end{cases} \\\\\displaystyle \begin{cases}\displaystyle -2 = -2 \\\\ \displaystyle\frac{1}{4} = \frac{1}{4}\displaystyle \end{cases} \\\\ - - - - - - - -\\\\\displaystyle \begin{cases}\displaystyle \frac{1}{x+y} = -\frac{1}{2} \\\displaystyle \frac{1}{x-y} = \frac{1}{4} \displaystyle \end{cases} \\\\\displaystyle x = 1 \;\; , \; \; y = -3 \\\\----------\\\\ \displaystyle \begin{cases}\displaystyle \frac{1}{1-3} - \frac{6}{1-(-3)} = -2 \\\\\displaystyle \frac{1}{1-3} + \frac{3}{1-(-3)} = \frac{1}{4} \displaystyle \end{cases} \\\\\displaystyle \begin{cases}\displaystyle -2 = -2 \\\\ \displaystyle\frac{1}{4} = \frac{1}{4}\displaystyle \end{cases} \\\\\\----------\\\\\\\displaystyle \huge \boxed{x=1 \; \; , \; \; y=-3}$