\(1,ĐKXĐ:x\ne\dfrac{3}{2}\\ \dfrac{2x+3}{2x-3}-\dfrac{3}{4x-6}=\dfrac{2}{5}\\ \Leftrightarrow\dfrac{2\left(2x+3\right)}{2\left(2x-3\right)}-\dfrac{3}{2\left(2x-3\right)}=\dfrac{2}{5}\\ \Leftrightarrow\dfrac{4x+6-3}{2\left(2x-3\right)}=\dfrac{2}{5}\\ \Rightarrow2\left(2x-3\right).2=5\left(4x+3\right)\\ \Leftrightarrow8x-12=20x+15\\ \Leftrightarrow12x+27=0\\ \Leftrightarrow x=-\dfrac{9}{4}\)
\(2,ĐKXĐ:x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12-\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \Rightarrow x^2+3x+2-5x+10=12-x^2+4\\ \Leftrightarrow x^2-2x+12=-x^2+16\\ \Leftrightarrow2x^2-2x-4=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x^2+x\right)-\left(2x+2\right)=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x-1}-\dfrac{7}{x-2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\\ \Leftrightarrow\dfrac{x-2-7\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{-1}{\left(x-1\right)\left(x-2\right)}\\ \Rightarrow x-2-7x+7+1=0\\ \Leftrightarrow-6x+6=0\\ \Leftrightarrow x=1\left(ktm\right)\)
2, đk : x khác -2 ; 2
\(\Rightarrow x^2+3x+2-5x+10=12+x^2-4\Leftrightarrow-2x+12=8\Leftrightarrow x=2\)(ktm)
-> vậy pt vô nghiệm
3, đk : x khác 1 ; 2
\(\Rightarrow x-2-7x+7=-1\Leftrightarrow-6x=-9\Leftrightarrow x=\dfrac{3}{2}\)(tm)
