Giải pt √2-x +√x +5√2x-x^2=7

Đáp án:

$S = \left\{ {0;2} \right\}$.

Giải thích các bước giải:

$\begin{array}{l} \,\,\,\,\,\sqrt {2 - x}  + \sqrt x  + 5\sqrt {2x - {x^2}}  = 7\,\,\left( 1 \right)\,\,\left( {0 \le x \le 2} \right)\\  \Leftrightarrow \sqrt {2 - x}  + \sqrt x  + 5\sqrt {x\left( {2 - x} \right)}  = 7\\ Dat\,\,t = \sqrt {2 - x}  + \sqrt x \,\,\left( {t \ge 0} \right)\\ \,\, \Rightarrow {t^2} = 2 - x + x + 2\sqrt {x\left( {2 - x} \right)} \\ \,\, \Rightarrow {t^2} = 2 + 2\sqrt {x\left( {2 - x} \right)} \\ \,\, \Rightarrow \sqrt {x\left( {2 - x} \right)}  = \frac{{{t^2} - 2}}{2}\\ \left( 1 \right) \Leftrightarrow t + 5.\frac{{{t^2} - 2}}{2} = 14\\ \,\,\,\,\,\, \Leftrightarrow 2t + 5{t^2} - 10 - 14 = 0\\  \Leftrightarrow 5{t^2} + 2t - 24 = 0\\  \Leftrightarrow \left[ \begin{array}{l} t = 2\,\,\,\,\,\,\,\,\,\,\left( {tm} \right)\\ t =  - \frac{{12}}{5}\,\,\,\left( {ktm} \right) \end{array} \right.\\ t = 2 \Rightarrow \sqrt {2 - x}  + \sqrt x  = 2\\ \,\,\,\,\,\,\,\,\,\, \Leftrightarrow 2 - x + x + 2\sqrt {x\left( {2 - x} \right)}  = 2\\ \,\,\,\,\,\,\,\,\, \Leftrightarrow 2\sqrt {x\left( {2 - x} \right)}  = 0\\ \,\,\,\,\,\,\,\,\, \Leftrightarrow x\left( {2 - x} \right) = 0\\ \,\,\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 2 \end{array} \right.\,\,\left( {tm} \right) \end{array}$

Vậy $S = \left\{ {0;2} \right\}$.