\(a.\dfrac{3x^2-3}{x^2-9}=\dfrac{1}{x-3}-\dfrac{4}{x+3}.\left(x\ne\pm3\right).\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{4}{x+3}-\dfrac{3x^2-3}{\left(x-3\right)\left(x+3\right)}=0.\)
\(\Leftrightarrow\dfrac{x+3-4x+12-3x^2+3}{\left(x-3\right)\left(x+3\right)}=0.\)
\(\Rightarrow-3x^2-3x+18=0.\Leftrightarrow\left(x-2\right)\left(x+3\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right).\\x=-3\left(koTM\right).\end{matrix}\right.\)
\(b.\left|3x-1\right|=2x-5.\Leftrightarrow\left\{{}\begin{matrix}2x-5\ge0.\\\left(3x-1\right)^2=\left(2x-5\right)^2.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}.\\9x^2-6x+1=4x^2-20x+25.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}.\\5x^2+14x-24=0.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}.\\\left(x-\dfrac{6}{5}\right)\left(x+4\right)=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}.\\\left[{}\begin{matrix}x=\dfrac{6}{5}\left(koTM\right).\\x=-4\left(koTM\right).\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\phi.\)
\(c.\left|2x+1\right|=\left|4x-7\right|.\Leftrightarrow\left(2x+1\right)^2=\left(4x-7\right)^2.\)
\(\Leftrightarrow4x^2+4x+1-16x^2+56x-49=0.\Leftrightarrow-12x^2+60x-48=0.\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=4.\\x=1.\end{matrix}\right.\)
\(d.\sqrt{3x^2-4x-4}=\sqrt{2x+5}.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+5\ge0.\\3x^2-4x-4=2x+5.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-5}{2}.\\3x^2-6x-9=0.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-5}{2}.\\\left[{}\begin{matrix}x=3.\\x=-1.\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=-1.\end{matrix}\right.\)
\(e.\sqrt{-2x^2-10x+9}=x+2.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2\ge\\-2x^2-10x+9=\left(x+2\right)^2.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2.\\-2x^2-10x+9-x^2-4x-4=0.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2.\\-3x^2-14x+5=0.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2.\\\left(x-\dfrac{1}{3}\right)\left(x+5\right)=0.\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1}{3}\left(TM\right).\)
