Question

Срооочнооооо
√х=4-х
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Answer 1

$\sqrt{x}=4-x\; \; \; \Leftrightarrow \; \; \; \left\{\begin{array}{l}x=(4-x)^2\\4-x\geq 0\end{array}\right\; \; \left\{\begin{array}{l}x=16-8x+x^2\\x\leq 4\end{array}\right\\\\\\\left\{\begin{array}{l}x^2-9x+16=0\\x\leq 4\end{array}\right\; \; \left\{\begin{array}{l}x_{1,2}=\frac{9\pm \sqrt{17}}{2}\\x\leq 4\end{array}\right\; \; \; \Rightarrow \; \; \; x=\frac{9-\sqrt{17}}{2}\\\\\\\Big[\; 0<\frac{9-\sqrt{17}}{2}<4\; \; ,\; \; \; \frac{9+\sqrt{17}}{2}>4\; \Big]\\\\Otvet:\; \; x=\frac{9-\sqrt{17}}{2}\; .$