`***`Lời giải`***`

2)

`2x-1-x^2`

`=-(x^2-2x+1)`

`=-(x-1)^2`

`->`Chọn B

3)

`x.(x – y ) + y.(x – y )`

`=(x-y)(x+y)`

`=x^2-y^2`

`->`Chọn B

4)

`(2x^3 y +6x^ 2 y ^2 – 4xy^2):2xy `
`=(2x^3y :2xy)+(6x^2y^2 :2xy)– (4xy^2:2xy) `
`=x^2 +3xy–2y `

`->`Chọn C

II/

1)
a)

`x^2-y^2-2x+2y`

`=(x-y)(x+y)-2(x-y)`

`=(x-y)(x+y-2)`

b)

`x^2-6x-7`

`=x^2+x-7x-7`

`=x(x+1)-7(x+1)`

`=(x+1)(x-7)`

2)

`(2x + 3) ^2 – 4(3 + x)(x – 3) = 21`

`<=>4x^2+12x+9 – 4(x^2-9) = 21`

`<=>4x^2+12x+9 – 4x^2+36 = 21`

`<=>12x+24=0 `

`<=>12(x+2)=0 `

`<=>x+2=0 `

`<=>x=-2`

Vậy `S={-2}`

3)

Thực hiện phép chia`(3x^3 + 10x^2 + m – 5):(3x+1)`

Ta được: `(3x^3 + 10x^2 + m – 5):(3x+1)=x^2+3x` dư `-3x+m-5`

Để `(3x^3 + 10x^2 + m – 5) ⋮ (3x+1)`

Thì `-3x+m-5=0`

`<=>m=3x+5`

Vậy `m=3x+5`

4)

`9x^2-12x+27`

`=9x^2-12x+4+23`

`=(3x-2)^2+23`

Với `∀x` Ta có: `(3x-2)^2≥0<=>(3x-2)^2+23≥23>0`

Vậy `9x^2-12x+27>0` với `∀x`

đáp án:

2, c

3,b

4,a

giải thích:

2, 2x – 1 – x²=-(2x+1+x²)=-(x+1)²

3, x.(x – y ) + y.(x – y )=(x+y)(x-y)=x²-y²

4, (2x³y +6x²y² – 4xy²) : 2xy=x²+3xy-2y

II-TL:

1,

a) x² – y² –2x + 2y=a) x² – y² –2x + 2y +1-1

                             =(x²-2x+1)-(y²-2y+1)

                             =(2x+1)²-(2y+1)²

                             =(2x+1-2y-1)(2x+1+2y+1)

                             =4(x-y)(x+y+1)

b) x²– 6x -7=x²+x-7x-7

                  =x(x+1)-7(x+1)

                  =(x-7)(x+1)