Bài `1:`
`a)`
`VT=(5-x)/(x^2-1)=(-(x-5))/(x^2-1)`
`=-(x-5)/(x^2-1)=(x-5)/(1-x^2)= VP`
`to đpcm`
`b)`
`VT=(6x^2-x-2)/(3x-2)=(6x^2+3x-4x-2)/(3x-2)`
`=(3x(2x+1)-2(2x+1))/(3x-2)=((3x-2)(2x+1))/(3x-2)`
`=2x+1`
`VP=(2x^2-5x-3)/(x-3)=(2x^2+x-6x-3)/(x-3)`
`=(x(2x+1)-3(2x+1))/(x-3)=((x-3)(2x+1))/(x-3)`
`=2x+1`
`to VT=VP`
`to đpcm`
Bài `2:`
`(3x+1)/(12x^2+x-1)=(x+3)/A`
`<=> (3x+1)/(12x^2+4x-3x-1)=(x+3)/A`
`<=> (3x+1)/([4x(3x+1)-(3x+1)])=(x+3)/A`
`<=> (3x+1)/((4x-1)(3x+1))=(x+3)/A`
`<=> 1/(4x-1)=(x+3)/A`
`<=> A=(x+3)(4x-1)`
`<=> A=4x^2-x+12x-3`
`<=> A=4x^2+11x-3`
Vậy đa thức `A` trong đẳng thức là : `4x^2+11x-3`
