Question

Tìm x:  \(\left(2x+1\right)^2-3\left(2-x\right)^2+2\left(x-1\right)\left(x+2\right)=4-2\left(1-x\right)\)

Answer 1

\(\Leftrightarrow4x^2+4x+1-3\left(x^2-4x+4\right)+2\left(x^2+x-2\right)=4-2+2x\)

\(\Leftrightarrow4x^2+4x+1-3x^2+12x-12+2x^2+2x-4=2x+2\)

\(\Leftrightarrow3x^2+18x-15-2x-2=0\)

\(\Leftrightarrow3x^2+16x-17=0\)

\(\text{Δ}=16^2-4\cdot3\cdot\left(-17\right)=460>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-16-2\sqrt{115}}{6}=\dfrac{-8-\sqrt{115}}{3}\\x_2=\dfrac{-8+\sqrt{115}}{3}\end{matrix}\right.\)