решить систему[tex]x + y + \sqrt{x + y} = 20 \\ x^{2} + y {}^{2} = 136 [/tex]. Created by пахан49. algebra-ru

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$\left \{ {{x+y+\sqrt{x+y}=20} \atop {x^2+y^2=136}} \right. \\\\a)\; \; x+y+\sqrt{x+y}=20\; ,\; \; \; t=\sqrt{x+y} \geq 0\; ,\; \; x+y \geq 0\; \to \; y \geq -x\\\\t^2+t-20=0\; ,\; \; t_1=-5\ \textless \ 0\; ,\; \; t_2=4\ \textgreater \ 0\; (!)\\\\\sqrt{x+y}=4\; ,\; \; x+y=16\; ,\; \; y=16-x\\\\b)\; \; x^2+y^2=136\\\\x^2+(16-x)^2=136\; ,\; \; 2x^2-32x+256=136\, \Big |:2\; ,\\\\x^2-16x+6x=0\; \to \; \; x_1=6\; ,\; \; x_2=10\\\\y=16-x\; ,\; y_1=10\; ,\; \; y_2=6\\\\Otvet:\; \; (6,10)\; ,\; \; (10,6)\; .$

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