Đáp án:
`S=\{2\}`
Giải thích các bước giải:
`ĐKXĐ:x\ne 1;x\ne -1`
`\frac{6}{(x-1)(x+1)}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}`
`⇔\frac{6}{(x-1)(x+1)}+5=\frac{8x-1}{4x+4}+\frac{12x-1}{4x-4}`
`⇔\frac{6}{(x-1)(x+1)}+\frac{5(x-1)(x+1)}{(x-1)(x+1)}=\frac{8x-1}{4(x+1)}+\frac{12x-1}{4(x-1)}`
`⇔\frac{6+5(x^2-1)}{(x-1)(x+1)}=\frac{(8x-1)(x-1)+(12x-1)(x+1)}{4(x+1)(x-1)}`
`⇔\frac{6+5x^2-5}{(x-1)(x+1)}=\frac{8x^2-8x-x+1+12x^2+12x-x-1}{4(x+1)(x-1)}`
`⇔\frac{5x^2+1}{(x-1)(x+1)}=(8x^2-9x+1+12x^2+11x-1)/(4(x+1)(x-1))`
`⇔\frac{4(5x^2+1)}{4(x-1)(x+1)}=\frac{20x^2+2x}{4(x+1)(x-1)}`
`⇔\frac{20x^2+4}{4(x-1)(x+1)}=\frac{20x^2+2x}{4(x+1)(x-1)}`
`⇒20x^2+4=20x^2+2x`
`⇔20x^2+2x-20x^2-4=0`
`⇔2x-4=0`
`⇔2x=4`
`⇔x=2(TM)`
Vậy `S=\{2\}`
