\(\frac{x}{x+1}-\frac{x-5}{x+6}=\frac{-55}{x^2+7x+6}\)
\(\Leftrightarrow\frac{x\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}-\frac{\left(x-5\right)\left(x+1\right)}{\left(x+1\right)\left(x+6\right)}=\frac{-55}{\left(x+1\right)\left(x+6\right)}\) \(\left(ĐKXĐ:x\ne-1;x\ne-6\right)\)
\(\Leftrightarrow x^2+6x-x^2+4x+5+55=0\)
\(\Leftrightarrow10x+60=0\)
\(\Leftrightarrow x=-6\)
Vậy phương trình có nghiệm \(x=-6.\)
\(\frac{x}{x+1}-\frac{x-5}{x+6}=-\frac{55}{x^2+7x+6}ĐKXĐ:x\ne-1;-6\)
\(\frac{x}{x+1}-\frac{x-5}{x+6}=-\frac{55}{\left(x+1\right)\left(x+6\right)}\)
\(x\left(x+6\right)-\left(x-5\right)\left(x+1\right)=-55\)
\(10x+5=-55\)
\(10x=-60\)
\(x=-6\)
Theo ĐKXĐ => pt vô nghiệm
