f/
\(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3x-2\right)}{3}+\dfrac{15\left(3x-1\right)}{3}=\dfrac{2\left(2x+1\right)\left(3x+1\right)}{3}+\dfrac{6x\left(3x+1\right)}{3}\)
\(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3x-2\right)}{3}+\dfrac{15\left(3x-1\right)}{3}-\dfrac{2\left(2x+1\right)\left(3x+1\right)}{3}-\dfrac{6x\left(3x+1\right)}{3}=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-2\right)+15\left(3x-1\right)-2\left(2x+1\right)\left(3x+1\right)-6x\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[3x-2+15-2\left(x+1\right)-6x\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-2+15-2x-2-6x\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(11-5x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+1=0\\11-5x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{11}{5}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S\in\left\{-\dfrac{1}{3};\dfrac{11}{5}\right\}\)
g/ \(\Leftrightarrow\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+2\right)^2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{6\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\left(3x+2\right)^2-6\left(3x-2\right)-9x^2=0\)
\(\Leftrightarrow9x^2+12x+4-18x+12-9x^2=0\)
\(\Leftrightarrow16-6x=0\)
\(\Leftrightarrow6x=16\Leftrightarrow x=\dfrac{16}{6}=\dfrac{8}{3}\)
pt có tập nghiệm \(S\in\left\{\dfrac{8}{3}\right\}\)
h/ \(\Leftrightarrow\dfrac{5x}{x+1}-\dfrac{3}{x-1}=\dfrac{5\left(x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{5x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{5\left(x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow5x\left(x-1\right)-3\left(x+1\right)-5\left(x^2+1\right)=0\)
\(\Leftrightarrow5x^2-5x-3x-3-5x^2-5=0\)
\(\Leftrightarrow-8x-8=0\)
\(\Leftrightarrow x=-1\)
Vậy pt có tập nghiệm \(S\in\left\{-1\right\}\)
