Đáp án+Giải thích các bước giải:
`b)ĐKXĐ:x\ne 2;x\ne -2`
`(x-1)/(x+2)-x/(x-2)=(5x-2)/(4-x^2)`
`⇔(x-1)/(x+2)-x/(x-2)=(2-5x)/(x^2-4)`
`⇔((x-1)(x-2))/((x+2)(x-2))-(x(x+2))/((x+2)(x-2))=(2-5x)/((x+2)(x-2))`
`⇒(x-1)(x-2)-x(x+2)=2-5x`
`⇔x^2-3x+2-x^2-2x=2-5x`
`⇔-5x+2=2-5x`
`⇔-5x+5x=2-2`
`⇔0x=0`
Vậy phương trình đã cho có vô số nghiệm`(x\ne 2;x\ne -2)`
`d)ĐKXĐ:x\ne 1;x\ne -1`
`6/(x^2-1)+5=(8x-1)/(4x+4)-(12x-1)/(4-4x)`
`⇔6/(x^2-1)+(5(x^2-1))/(x^2-1)=(8x-1)/(4(x+1))-(12x-1)/(4(1-x))`
`⇔(6+5(x^2-1))/(x^2-1)=(8x-1)/(4(x+1))+(12x-1)/(4(x-1))`
`⇔(6+5x^2-5)/(x^2-1)=((x-1)(8x-1))/(4(x+1)(x-1))+((x+1)(12x-1))/(4(x+1)(x-1))`
`⇔(5x^2+1)/(x^2-1)=(8x^2-9x+1+12x^2+11x-1)/(4(x+1)(x-1))`
`⇔(5x^2+1)/(x^2-1)=(20x^2+2x)/(4(x+1)(x-1))`
`⇔(4(5x^2+1))/(4(x^2-1))=(20x^2+2x)/(4(x+1)(x-1))`
`⇒4(5x^2+1)=20x^2+2x`
`⇔20x^2+4=20x^2+2x`
`⇔20x^2-20x^2-2x=-4`
`⇔-2x=-4`
`⇔x=2(TM)`
Vậy `S=\{2\}`
