Question

Giải phương trình 2√(3x-2)=x^2+x-2

Answer 1

ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(2\sqrt{3x-2}-4=x^2+x-6\)

\(\Leftrightarrow\dfrac{6\left(x-2\right)}{\sqrt{3x-2}+4}=\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{6}{\sqrt{3x-2}+4}=x+3\left(1\right)\end{matrix}\right.\)

Xét (1), do \(x\ge\dfrac{2}{3}\Rightarrow VP=x+3>3\)

\(VT=\dfrac{6}{\sqrt{3x-2}+4}\le\dfrac{6}{4}< 3\)

\(\Rightarrow\left(1\right)\) vô nghiệm