Question

Answer 1

\(2x.\left(8x-1\right)^2.\left(4x-1\right)=9\)

\(\Leftrightarrow\left(64x^2-16x+1\right).\left(8x^2-2x\right)=9\)

Đặt \(8x^2-2x=t\)

\(\Rightarrow64x^2-16x=8t\)

Pt trở thành:

\(\left(8t+1\right).t=9\)

\(\Leftrightarrow8t^2+t-9=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{9}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}8x^2-2x=1\\8x^2-2x=-\dfrac{9}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}8x^2-2x-1=0\\64x^2-16x+9=0\end{matrix}\right.\)

\(\Leftrightarrow...\)