Question

Решите систему уравнений​

Answer 1

$\left\{\begin{array}{l}(x-y)^2\cdot |x+y|=24\\\frac{(x+y)^2}{|x-y|}=18\end{array}\right\; x\ne y,\; (x-y)^2=|x-y|^2\; ,\; (x+y)^2=|x+y|^2\\\\\\\left\{\begin{array}{l}|x-y|^2\cdot |x+y|=24\; ,\\\frac{|x+y|^2}{|x-y|}=18\; ,\end{array}\right\; \; \; u=|x-y|\; ,\; v=|x+y|\; \; \; \left\{\begin{array}{l}u^2\cdot v=24\\\frac{v^2}{u}=18\end{array}\right$

$\left\{\begin{array}{l}u^2v=24\\v^2=18u\end{array}\right\; \; \left\{\begin{array}{l}v=\frac{24}{u^2}\\\frac{24^2}{u^4}=18u}\end{array}\right\; \; \left\{\begin{array}{l}v=\frac{24}{u^2}\\u^5=32\end{array}\right\; \; \left\{\begin{array}{l}v=6\\u=2\end{array}\right\; \; \left\{\begin{array}{l}|x+y|=6\\|x-y|=2\end{array}\right$

$a)\left\{\begin{array}{l}x+y=-6\\x-y=-2\end{array}\right\; b)\left\{\begin{array}{l}x+y=-6\\x-y=2\end{array}\right\; c)\left\{\begin{array}{l}x+y=6\\x-y=-2\end{array}\right\; d)\left\{\begin{array}{l}x+y=6\\x-y=2\end{array}\right$

$a)\left\{\begin{array}{l}2x=-6-2\\2y=-6+2\end{array}\right\; \; \left\{\begin{array}{l}x=\frac{-6-2}{2}\\y=\frac{-6+2}{2}\end{array}\right\; \; \left\{\begin{array}{l}x=-4\\y=-2\end{array}\right$

$b)\left\{\begin{array}{l}2x=-6+2\\2y=-6-2\end{array}\right\; \; \left\{\begin{array}{l}x=\frac{-6+2}{2}\\y=\frac{-6-2}{2}\end{array}\right\; \; \left\{\begin{array}{ccc}x=-2\\y=-4\end{array}\right$

$c)\left\{\begin{array}{l}2x=6-2\\2y=6+2\end{array}\right\; \; \left\{\begin{array}{l}x=2\\y=4\end{array}\right$

$d)\left\{\begin{array}{l}2x=6+2\\2y=6-2\end{array}\right\; \; \left\{\begin{array}{l}x=4\\y=2\end{array}\right$

$Otvet:\; (-4,-2)\; ,\; (-2,-4)\; ,\; (2,4)\; ,\; (4,2)\; .$