Question

giải ác phương trình sau:

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

2)\(\frac{x}{x-1}-\frac{5x-3}{x^2-1}=0\)

3)\(\frac{1}{x-3}-\frac{4}{x+3}=\frac{3x}{9-x^2}\)

4)\(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)

5)\(\frac{-3}{2x}-\frac{x+1}{x+2}=\frac{-3}{x\left(x+2\right)}\)

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

Answer 1

1, Đk x≠2;-2

\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}=0\\ =>\frac{x+2}{2\left(x-2\right)}-\frac{4x}{\left(x-2\right).\left(x+2\right)}=0\\ =>\frac{\left(x+2\right)^2}{2\left(x^2-4\right)}-\frac{8x}{2\left(x-2\right).\left(x+2\right)}=0\\ =>\frac{x^2+4x+4-8x}{2\left(x-2\right)\left(x+2\right)}=0\\ =>\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}=0\\ =>\frac{x-2}{2\left(x+2\right)}=0\\ =>x-2=0\\ =>x=2\left(loại\right)\)