a: \(\dfrac{x^2-1}{3}=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=6\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)
=>x=1 hoặc x=5
b: \(\dfrac{3}{x-2}+\dfrac{7}{x+2}=\dfrac{8x}{x^2-4}\)
=>3x+6+7x-4=8x
=>10x+2=8x
=>2x=-2
hay x=-1