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 \(S = \left\{ {0;2} \right\}\).