a: \(\Leftrightarrow x-2-x-2=3x-12\)
=>3x-12=-4
=>3x=8
hay x=8/3(nhận)
b: Suy ra: \(\dfrac{3\left(-x^2+12x+4\right)}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36}{3\left(x+4\right)\left(x-1\right)}+\dfrac{x+4}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36+x+4\)
\(\Leftrightarrow-3x^2+35x-28=0\)
\(\Leftrightarrow3x^2-35x+28=0\)
\(\text{Δ}=\left(-35\right)^2-4\cdot3\cdot28=889>0\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{35-\sqrt{889}}{6}\\x_2=\dfrac{35+\sqrt{889}}{6}\end{matrix}\right.\)
c: \(\Leftrightarrow x^2+x+1+2x^2-5=4x-4\)
\(\Leftrightarrow3x^2+x-4x=0\)
=>3x(x-1)=0
=>x=0(nhận) hoặc x=1(loại)
