Question

2x² + 3x - 4 = (4x - 3)\(\sqrt{3x-1}\)

Answer 1

ĐKXĐ: \(x\ge\dfrac{1}{3}\)

Đặt \(\sqrt{3x-1}=t\ge0\Rightarrow3x-1=t^2\)

\(\Rightarrow\left\{{}\begin{matrix}2x^2+3x-4=\left(4x-3\right)t\\3x-1=t^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+3x-4=4tx-3t\\2t^2=6x-2\end{matrix}\right.\)

\(\Leftrightarrow2x^2+2t^2+3x-4=4tx-3t+6x-2\)

\(\Leftrightarrow2\left(x-t\right)^2-3\left(x-t\right)-2=0\)

\(\Leftrightarrow...\)