Question

[Nguyễn Hoàng Vũ]

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

Answer 1

[Nguyễn Lê Phước Thịnh]

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

\(\Leftrightarrow x^2-4x+4-2x^2-4x-2=\left(2x-6x^2+1-3x\right)+2x^2-2x\)

\(\Leftrightarrow-x^2-8x+2=-6x^2-x+1+2x^2-2x\)

\(\Leftrightarrow-x^2-8x+2=-4x^2-3x+1\)

\(\Leftrightarrow3x^2-5x+1=0\)

\(\Delta=\left(-5\right)^2-4\cdot3\cdot1=25-12=13>0\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{5-\sqrt{13}}{6}\\x_2=\dfrac{5+\sqrt{13}}{6}\end{matrix}\right.\)