Question

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

Answer 1

\(\Leftrightarrow x^2-6x+9-4x^2-4x-1-2\left(x^2+x-2\right)=3\left(x-3\right)-\left(4x^2+8x-x-2\right)\)

\(\Leftrightarrow-3x^2-10x+8-2x^2-2x+4=3\left(x-3\right)-4x^2-7x+2\)

\(\Leftrightarrow-5x^2-12x+12=3x-9-4x^2-7x+2\)

\(\Leftrightarrow-5x^2-12x+12=-4x^2-4x-7\)

\(\Leftrightarrow-4x^2-4x-7+5x^2+12x-12=0\)

\(\Leftrightarrow x^2+8x-19=0\)

\(\text{Δ}=8^2-4\cdot1\cdot\left(-19\right)=76+64=140\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-8-2\sqrt{35}}{2}=-4-\sqrt{35}\\x_2=-4+\sqrt{35}\end{matrix}\right.\)