giải phương trình a: 2 × √3x-1 – 4 = 4x b: √x+ √2x-1 + √x- √2x-1 = √2

Đáp án:

$\begin{array}{l}
a)2\sqrt {3x – 1}  – 4 = 4x\\
 \Rightarrow \sqrt {3x – 1}  = 2x + 2\left( {dk:x \ge \frac{1}{3}} \right)\\
 \Rightarrow 3x – 1 = {\left( {2x + 2} \right)^2}\\
 \Rightarrow 3x – 1 = 4{x^2} + 8x + 4\\
 \Rightarrow 4{x^2} + 5x + 5 = 0\\
 \Rightarrow 4{x^2} + 2.2x.\frac{5}{4} + \frac{{25}}{{16}} + \frac{{55}}{{16}} = 0\\
 \Rightarrow {\left( {2x + \frac{5}{4}} \right)^2} + \frac{{55}}{{16}} = 0\left( {vô\,nghiệm} \right)\\
 \Rightarrow x \in \emptyset \\
b)\\
\sqrt {x + \sqrt {2x – 1} }  + \sqrt {x – \sqrt {2x – 1} }  = \sqrt 2 \left( {dkxd:x \ge \frac{1}{2}} \right)\\
 \Rightarrow x + \sqrt {2x – 1}  + 2\sqrt {x + \sqrt {2x – 1} } .\sqrt {x – \sqrt {2x – 1} }  + x – \sqrt {2x – 1}  = 2\\
 \Rightarrow 2x + 2\sqrt {{x^2} – {{\left( {\sqrt {2x – 1} } \right)}^2}}  = 2\\
 \Rightarrow \sqrt {{x^2} – 2x + 1}  = 1 – x\left( {dk:x \le 1} \right)\\
 \Rightarrow {x^2} – 2x + 1 = {x^2} – 2x + 1\left( {\forall x} \right)\\
 \Rightarrow \frac{1}{2} \le x \le 1
\end{array}$