`c, (x+1)/(x-1) - (x-1)/(x+1)=4/(x^2-1)` ; ĐKXĐ : `x` $\neq$ `+-1`
`<=>((x+1)^2-(x-1)^2)/((x-1)(x+1)` `=``4/((x-1)(x+1))`
`<=>(x^2+2x+1-(x^2-2x+1))/((x-1)(x+1))``=``4/((x-1)(x+1))`
`<=>(x^2+2x+1-x^2+2x-1)/((x-1)(x+1))``=``4/((x-1)(x+1))`
`=>x^2+2x+1-x^2+2x-1=4`
`<=>4x=4`
`<=>x=1` ( không thỏa mãn đkxđ )
Vậy phương trình trên vô nghiệm
`d, 2x- (2x^2)/(x+3)= (4x)/(x+3) + 2/7` ; ĐKXĐ : `x` $\neq$ `-3`
`<=>((2x.7)(x+3)-(2x^2. 7))/(7(x+3))=((4x. 7)+2(x+3))/(7(x+3))`
`<=>(14x(x+3)-14x^2)/(7(x+3))=(28x+2x+6)/(7(x+3))`
`=>14x^2+42x-14x^2=28x+2x+6`
`<=>14x^2-14x^2+42x-28x-2x=6`
`<=>12x=6`
`<=>x=1/2`
Vậy `S={1/2}`
