Question

Giải phương trình (kiểu lớp 10)

\(\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)

Answer 1

\(\Leftrightarrow3\left(2x^2+1\right)+\left(8x-3\right)\sqrt{2x^2+1}-3x^2+x=0\)

Đặt \(\sqrt{2x^2+1}=t>0\)

\(\Rightarrow3t^2+\left(8x-3\right)t-3x^2+x=0\)

\(\Delta=\left(8x-3\right)^2-12\left(-3x^2+x\right)=\left(10x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-8x+3-\left(10x-3\right)}{6}=-3x+1\\t=\dfrac{-8x+3+10x-3}{6}=\dfrac{x}{3}\end{matrix}\right.\)

\(\Leftrightarrow...\)