Question

Tìm nghiệm của pt sau:

\(\frac{x}{x+1}-\frac{x-5}{x+6}=\frac{-55}{x^2+7x+6}\)

Answer 1

https://i.imgur.com/8mTs6Xw.jpg

Answer 2

\(\frac{x}{x+1}-\frac{x-5}{x+6}=\frac{-55}{x^2+7x+6}\)

\(\Leftrightarrow\frac{x\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}-\frac{\left(x-5\right)\left(x+1\right)}{\left(x+1\right)\left(x+6\right)}=\frac{-55}{\left(x+1\right)\left(x+6\right)}\) \(\left(ĐKXĐ:x\ne-1;x\ne-6\right)\)

\(\Leftrightarrow x^2+6x-x^2+4x+5+55=0\)

\(\Leftrightarrow10x+60=0\)

\(\Leftrightarrow x=-6\)

Vậy phương trình có nghiệm \(x=-6.\)

Answer 3

\(\frac{x}{x+1}-\frac{x-5}{x+6}=-\frac{55}{x^2+7x+6}ĐKXĐ:x\ne-1;-6\)

\(\frac{x}{x+1}-\frac{x-5}{x+6}=-\frac{55}{\left(x+1\right)\left(x+6\right)}\)

\(x\left(x+6\right)-\left(x-5\right)\left(x+1\right)=-55\)

\(10x+5=-55\)

\(10x=-60\)

\(x=-6\)

Theo ĐKXĐ => pt vô nghiệm