Question

Решить систему уравнений

Answer 1

$\left\{\begin{array}{l}x^2y+xy^2=30\\\frac{1}{x}+\frac{1}{y}=\frac{5}{6}\end{array}\right\; \; \left\{\begin{array}{l}xy(x+y)=30\\\frac{x+y}{xy}=\frac{5}{6}\end{array}\right\; \; \; \; u=x+y\; ,\; v=xy\ne 0\\\\\\\left\{\begin{array}{ccc}uv=30\\\frac{u}{v}=\frac{5}{6} \end{array}\right\; \; \left\{\begin{array}{l}uv=30\\v=\frac{6u}{5}\end{array}\right\; \; \left\{\begin{array}{l}\frac{6u^2}{5}=30\\v=\frac{6u}{5}\end{array}\right\; \; \left\{\begin{array}{l}u^2=25\\v=\frac{6u}{5}\end{array}\right$

$a)\; \; \left\{\begin{array}{l}u=x+y=-5\\v=xy=-6\end{array}\right\; \; \left\{\begin{array}{l}y=-x-5\\x(-x-5)=-6\end{array}\right\; \; \left\{\begin{array}{l}y=-x-5\\x^2+5x-6=0\end{array}\right\\\\\\\left\{\begin{array}{ccc}y_1=1\; ,\; y_2=-6\\x_1=-6\; ,\; x_2=1\end{array}\right$

$b)\; \; \left\{\begin{array}{l}u=x+y=5\\v=xy=6\end{array}\right\; \; \left\{\begin{array}{l}y=-x+5\\x(-x+5)=6\end{array}\right\; \; \left\{\begin{array}{l}y=-x+5\\x^2-5x+6=0\end{array}\right\\\\\\\left\{\begin{array}{ccc}y_1=3\; ,\; y_2=2\\x_1=2\; ,\; x_2=3\end{array}\right\\\\\\Otvet:\; \; (-6;1)\; ,\; (1;-6)\; ,\; (2;3)\; ,\; (3;2)\; .$