Question

Giai pt:

\(\sqrt{2-3x}=-3x^2+7x-1\)

Answer 1

Lời giải:

ĐKXĐ: \(x\leq \frac{2}{3}\)

Ta có: \(\sqrt{2-3x}=-3x^2+7x-1\)

\(\Leftrightarrow 3x^2-7x+1+\sqrt{2-3x}=0\)

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

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

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

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

Tức là \(x-2-\frac{1}{\sqrt{2-3x}+1}\neq 0\Rightarrow 3x-1=0\Rightarrow x=\frac{1}{3}\) (t/m)

Vậy...........