Tìm các đa thức A, B, C trong các đẳng thức sau:

Đáp án+Giải thích các bước giải:

 `a)`

`\frac{3x^{2}-2x-5}{A}=\frac{3x-5}{2x-3}`

`=>A.(3x-5)=(3x^{2}-5x+3x-5).(2x-3)`

`=>A=\frac{[3x.(x+1)-5.(x+1)].(2x-3)}{3x-5}`

`=>A=\frac{(3x-5).(x+1).(2x-3)}{3x-5}`

`=>A=(x+1).(2x-3)`

`=>A=2x^{2}-3x+2x-3`

`=>A=2x^{2}-x-3`

Vậy `A=2x^{2}-x-3`

`b)`

`\frac{2x^{2}+3x-2}{x^{2}-4}=\frac{B}{x^{2}-4x+4}`

`=>B.(x^{2}-2^{2})=(2x^{2}+4x-x-2).(x^{2}-2.x.2+2^{2})`

`=>B=\frac{[2x.(x+2)-1.(x+2)].(x-2)^{2}}{(x-2).(x+2)}`

`=>B=\frac{(2x-1).(x+2).(x-2)}{x+2}`

`=>B=(2x-1).(x-2)`

`=>B=2x^{2}-4x-x+2`

`=>B=2x^{2}-5x+2`

Vậy `B=2x^{2}-5x+2`

`c)`

`\frac{C}{2x+1}=\frac{10x^{2}-5x}{4x^{2}-1}`

`=>C.[(2x)^{2}-1^{2}]=(2x+1).5x.(2x-1)`

`=>C=\frac{(2x+1).(2x-1).5x}{(2x-1).(2x+1)}`

`=>C=5x`

Vậy `C=5x`

`d)`

`\frac{4x^{2}-16x+16}{x^{2}-4}=\frac{D}{x+2}`

`=>D.(x^{2}-2^{2})=4.(x^{2}-4x+4).(x+2)`

`=>D=\frac{4.(x^{2}-2.x.2+2^{2}).(x+2)}{(x-2).(x+2)}`

`=>D=\frac{4.(x-2)^{2}}{x-2}`

`=>D=4.(x-2)`

`=>D=4x-8`

Vậy `D=4x-8`