ĐK: \(x\ne-2;-3;-4;-5\)
\(1+\dfrac{1}{x+2}-\left(1+\dfrac{1}{x+3}\right)=1+\dfrac{1}{x+4}-\left(1+\dfrac{1}{x+5}\right)\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=\left(x+4\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+5x+6=x^2+9x+20\)
\(\Leftrightarrow4x=-14\Rightarrow x=-\dfrac{7}{2}\)
b/ ĐK: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{x+7}{x+2}-\dfrac{12}{x^2-4}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}-\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}-\dfrac{12}{x^2-4}=0\)
\(\Leftrightarrow x^2+3x+2-\left(x^2+5x-14\right)-12=0\)
\(\Leftrightarrow-2x+4=0\Rightarrow x=2\) (ko t/m)
Vậy pt vô nghiệm