Question

Решите уравнение: √x+1-√9-x = √2x-12

Answer 1

$\sqrt{x+1}-\sqrt{9-x}=\sqrt{2x-12}\\\\(x+1)-2\sqrt{(x+1)(9-x)} +(9-x)=2x-12\\\\-2\, \sqrt{(x+1)(9-x)}=2x-22\\\\\sqrt{(x+1)(9-x)}=11-x\\\\9x-x^2+9-x=121-22x+x^2\\\\2x^2-30x+112=0\\\\x^2-15x+56=0\; \; \to \; \; \; x_1=7\; ,\; \; x_2=8\; \; (teorema\; Vieta)\\\\Proverka:\\\\x=7:\; \; \sqrt8-\sqrt2=\sqrt2\; ,\; \; 2\sqrt2-\sqrt2=\sqrt2\; ,\; \; \sqrt2=\sqrt2\; \; verno\\\\x=8:\; \; \sqrt9-\sqrt1=\sqrt4\; ,\; \; 3-1=2\; ,\; \; 2=2\; \; verno\\\\Otvet:\; \; x=7\; ,\; x=8\; .$

P.S.

Можно было написать ОДЗ.

$ODZ:\; \left\{\begin{array}{ccc}x+1\geq 0\\9-x\geq 0\\2x-12\geq 0\end{array}\right\; \; \left\{\begin{array}{ccc}x\geq -1\\x\leq 9\\x\geq 6\end{array}\right\; \; \left\{\begin{array}{cc}x\leq 9\\x\geq 6\end{array}\right\; \; \; \Rightarrow \; \; 6\leq x\leq 9$

Answer 2

Смотри.....................

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