Question

giải pt

\(\sqrt{3x+1}-\sqrt{2-x}=\frac{4x-1}{3}\)

Answer 1

ĐKXĐ: \(-\frac{1}{3}\le x\le2\)

\(\Leftrightarrow\frac{4x-1}{\sqrt{3x+1}+\sqrt{2-x}}-\frac{4x-1}{3}=0\)

\(\Leftrightarrow\left(4x-1\right)\left(\frac{1}{\sqrt{3x+1}+\sqrt{2-x}}-\frac{1}{3}\right)=0\)

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

\(\left(1\right)\Leftrightarrow2x+3+2\sqrt{\left(3x+1\right)\left(2-x\right)}=9\)

\(\Leftrightarrow\sqrt{-3x^2+5x+2}=3-x\)

\(\Leftrightarrow-3x^2+5x+2=x^2-6x+9\)

\(\Leftrightarrow4x^2-11x+7=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{7}{4}\end{matrix}\right.\)