Đáp án:
$\begin{array}{l}
a)\sqrt {2{x^2} - x + 3} - \sqrt {21x - 17} = x - {x^2}\\
Dkxd:\left\{ \begin{array}{l}
2{x^2} - x + 3 \ge 0\forall x\\
21x - 17 \ge 0
\end{array} \right. \Rightarrow x \ge \frac{{17}}{{21}}\\
Pt \Leftrightarrow \sqrt {2{x^2} - x + 3} - 2 + 2 - \sqrt {21x - 17} = x - {x^2}\\
\Rightarrow \frac{{2{x^2} - x + 3 - 4}}{{\sqrt {2{x^2} - x + 3} + 2}} + \frac{{4 - \left( {21x - 17} \right)}}{{2 + \sqrt {21x - 17} }} = x\left( {1 - x} \right)\\
\Rightarrow \frac{{2{x^2} - x - 1}}{{\sqrt {2{x^2} - x + 3} + 2}} + \frac{{21 - 21x}}{{2 + \sqrt {21x - 17} }} = x\left( {1 - x} \right)\\
\Rightarrow \frac{{\left( {x - 1} \right)\left( {2x + 1} \right)}}{{\sqrt {2{x^2} - x + 3} + 2}} + \frac{{21\left( {1 - x} \right)}}{{2 + \sqrt {21x - 17} }} = x\left( {1 - x} \right)\\
\Rightarrow \left( {1 - x} \right)\left( {x - \frac{{21}}{{2 + \sqrt {21x - 17} }} + \frac{{2x + 1}}{{\sqrt {2{x^2} - x + 3} + 2}}} \right) = 0\\
\Rightarrow 1 - x = 0\\
\left( {do:x - \frac{{21}}{{2 + \sqrt {21x - 17} }} + \frac{{2x + 1}}{{\sqrt {2{x^2} - x + 3} + 2}} > 0x\forall x \ge \frac{{17}}{{21}}} \right)\\
\Rightarrow x = 1\left( {tmdk} \right)
\end{array}$
