Question

Решать уравненеие) подкорень2-x+подкорень х+3=3

Answer 1

$\sqrt{2-x}+\sqrt{x+3}=3\\OD3.:\\\begin{cases}2-x\geq0\\x+3\geq0\end{cases}\Rightarrow\begin{cases}x\leq2\\x\geq-3\end{cases}\Rightarrow x\in[-3;\;2]\\\\(\sqrt{2-x}+\sqrt{x+3})^2=9\\2-x+2\sqrt{2-x}\sqrt{x+3}+x+3=9\\2\sqrt{(2-x)(x+3)}=4\\\sqrt{6-x-x^2}=2\\(\sqrt{6-x-x^2})^2=4\\6-x-x^2=4\\x^2-x-2=0\\D=1+4\cdot2=9\\x_{1,2}=\frac{1\pm3}2\\x_1=-1,\;x_2=2$