\(a,\\ \Leftrightarrow\left(x-2\right)-\left(x+2\right)=3x-12\\ \Leftrightarrow x-2-x-2=3x-12\\ \Leftrightarrow3x=8\Rightarrow x=\dfrac{8}{3}\\ b,\\ \Leftrightarrow\left(-x^2+12x+4\right)=12\left(x-1\right)+4\left(x+4\right)\\ \Leftrightarrow x^2+4x+4=12x-12+4x+16\\ \Leftrightarrow x\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
\(c,\\ \Leftrightarrow\left(-x^2+2x+1\right)+\left(2x^2-5\right)=4\left(x-1\right)\\ \Leftrightarrow3x^2-3x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
\(d,\\ \Leftrightarrow\dfrac{5}{x^2-2x+4}=\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}\\ \Leftrightarrow\left\{{}\begin{matrix}5=\left(x^2-2x+4\right)\left(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}\right)\\x^2-2x+4\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x=-4\\x^2-2x+4\ne0\end{matrix}\right.\)
Bài 2 :
a. \(\dfrac{1}{x+2}-\dfrac{1}{x-2}=\dfrac{3x-12}{x^2-4}\\ ĐKXĐ:x\ne-2,x\ne2\)
\(\Leftrightarrow\dfrac{1.\left(x-2\right)}{x^2-4}-\dfrac{1.\left(x+2\right)}{x^2-4}=\dfrac{3x+12}{x^2-4}\\ \Leftrightarrow x-2-x-2=3x+12\\ \Leftrightarrow x-x-3x=12+2+2\\ \Leftrightarrow-3x=18\\ \Leftrightarrow x=-6\left(nhận\right)\)
Vậy PT có tập nghiệm S = { 6 }
