Em chia nhỏ từng câu ra để hỏi nhé
1) \(\begin{array}{l}
3\left( {\sqrt x + \sqrt {1 - x} } \right) = 3 + 2\sqrt {x - {x^2}} \,\,\left( {DK:\,0 \le x \le 1} \right)\\
Dat\,\,\sqrt x + \sqrt {1 - x} = t\left( {t \ge 0} \right)\\
\Rightarrow {t^2} = x + 1 - x + 2\sqrt {x\left( {1 - x} \right)} \\
\Leftrightarrow {t^2} = 1 + 2\sqrt {x - {x^2}} \Leftrightarrow 2\sqrt {x - {x^2}} = {t^2} - 1\\
pt \Rightarrow 3t = 3 + {t^2} - 1 \Leftrightarrow {t^2} - 3t + 2 = 0\\
\Leftrightarrow \left( {t - 1} \right)\left( {t - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 1\left( {tm} \right)\\
t = 2\left( {tm} \right)
\end{array} \right.\\
+ )\,t = 1 \Rightarrow 2\sqrt {x - {x^2}} = {1^2} - 1\\
\Leftrightarrow x - {x^2} = 0\\
\Leftrightarrow x\left( {1 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\left( {tm} \right)\\
+ )t = 2 \Rightarrow \,2\sqrt {x - {x^2}} = {2^2} - 1\\
\Leftrightarrow 2\sqrt {x - {x^2}} = 3\\
\Leftrightarrow 4x - 4{x^2} = 9\\
\Leftrightarrow 4{x^2} - 4x + 9 = 0\left( {VN} \right)
\end{array}\)
