a) \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\left(x\ne-7;x\ne\dfrac{3}{2}\right)\)
\(\Rightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
\(\Leftrightarrow x=-\dfrac{1}{56}\) (tm)
b) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\left(x\ne\pm1\right)\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\dfrac{4}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\left(ktm\right)\)
Vậy pt vô nghiệm
