`***`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`