Question

( 2x² - 3x - 1 )² - 3(2x² - 3x - 5 ) = 16

Answer 1

Đặt \(t=2x^2-3x-1\)

\(\Rightarrow t^2-3\left(t-4\right)-16=0\)

\(\Rightarrow t^2-3t+12-16=0\)

\(\Rightarrow t^2-3t-4=0\)

\(\Rightarrow\left\{{}\begin{matrix}t_1=-1\\t_2=4\end{matrix}\right.\)

\(TH_1:t=-1\)

\(\Leftrightarrow2x^2-3x-1=-1\)

\(\Leftrightarrow2x^2-3x=0\)

\(\Leftrightarrow x\left(2x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(TH_2:t=4\)

\(\Leftrightarrow2x^2-3x-1=4\)

\(\Leftrightarrow2x^2-3x-5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=\dfrac{5}{2}\end{matrix}\right.\)