giải pt

1,\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)

2,\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)

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

4,\(\frac{2x}{x-1}+\frac{4}{x^2+2x-3=}=\frac{2x-5}{x+3}\)

5,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)

6,\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)

7,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)