Question

x-1/x+2-x+6/4-x²=x/x-2

Answer 1

điều kiện \(x\ne\pm2\)

\(\frac{x-1}{x-2}+\frac{x-6}{x^2-4}=\frac{x}{x-2}\)

\(\frac{x-1}{x-2}+\frac{x-6}{x^2-4}-\frac{x}{x-2}=0\)

\(\frac{\left(x-1\right)\left(x+2\right)+\left(x-6\right)-x\left(x+2\right)}{x^2-4}=0\)

\(x^2+x-2+x-6-x^2-2x=0\)

0x=8

phương trình vô nghiệm

Answer 2

PT: \(\frac{x-1}{x+2}-\frac{x+6}{4-x^2}=\frac{x}{x-2}\) \(\left(x\ne\pm2\right)\)

\(\Rightarrow\left(x-1\right)\left(x-2\right)+x+6=x\left(x+2\right)\)

\(\Leftrightarrow x^2-3x+2+x+6=x^2+2x\)

\(\Leftrightarrow-4x=-8\) \(\Leftrightarrow x=2\left(loại\right)\)

Vậy phương trình vô nghiệm