Question

Giải pt: \(x^2-x+1=2\sqrt{3x-1}\)

Answer 1

Lời giải:

ĐKXĐ: \(x\geq \frac{1}{3}\)

Ta có: \(x^2-x+1=2\sqrt{3x-1}\)

\(\Leftrightarrow x^2+2x+1=3x+2\sqrt{3x-1}\)

\(\Leftrightarrow (x+1)^2=(3x-1)+2\sqrt{3x-1}+1\)

\(\Leftrightarrow (x+1)^2=(\sqrt{3x-1}+1)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{3x-1}+1\left(1\right)\\x+1=-\sqrt{3x-1}-1\left(2\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow x=\sqrt{3x-1}\Leftrightarrow x^2=3x-1\) (do $x>0$)

\(\Leftrightarrow x^2-3x+1=0\Leftrightarrow x=\frac{3\pm \sqrt{5}}{2}\) (thỏa mãn)

(2) \(\Leftrightarrow x+\sqrt{3x-1}+2=0\)

Điều trên vô lý do

\(x\geq \frac{1}{3}; \sqrt{3x-1}\geq 0\Rightarrow x+\sqrt{3x-1}+2\geq 2+\frac{1}{3}>0\)

Vậy \(x=\frac{3\pm \sqrt{5}}{2}\)