Question

система иррациональных уравнений. 11 класс. хелпуем пасаны

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

$\sqrt{x+y}=a,\sqrt{x^2+xy}=b;a,b\geq 0\\ \left\{\begin{array}{c}a+b=3\\a^2+b^2=5\end{array}\right.=>\left\{\begin{array}{c}a+b=3\\(a+b)^2-(a^2+b^2)=3^2-5\end{array}\right.=>\\ \left\{\begin{array}{c}a+b=3\\ab=2\end{array}\right.=>\left\{\begin{array}{c}a=3-b\\(3-b)b=2\;(1)\end{array}\right.\\ (1)\;b^2-3b+2=0=>b=2\;\;\;\;\;\;\;\;\;b=1\\ =>\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; a=1\;\;\;\;\;\;\;\;\;a=2$

$[1] \left\{\begin{array}{c}\sqrt{x+y}=1\\\sqrt{x^2+xy}=2\end{array}\right.=>\left\{\begin{array}{c}x+y=1\\\sqrt{x}=2\end{array}\right.=>\left\{\begin{array}{c}x=4\\y=-3\end{array}\right.$

$[2] \left\{\begin{array}{c}\sqrt{x+y}=2\\\sqrt{x^2+xy}=1\end{array}\right.=>\left\{\begin{array}{c}x+y=4\\\sqrt{4x}=1\end{array}\right.=>\left\{\begin{array}{c}x=\dfrac{1}{4}\\y=3\dfrac{3}{4}\end{array}\right.$