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Bài `2.`
`a,x^2-4y^2=x^2-(2y)^2=(x-2y)(x+2y)`
`b, 3x(x-1)-2(1-x)=3x(x-1)+2(x-1)=(x-1)(3x+2)`
`c, x^2-y^2-4x+4=(x^2-4x+4)-y^2=(x-2)^2 -y^2=(x-2-y)(x-2+y)`
`d, x^2 - x - 2020.2021=x^2 - 2021x + 2020x - 2020.2021`
`= x (x-2021) + 2020 (x-2021)`
`= (x-2021)(x+2020)`
Bài `3.`
Đặt `f(x)=2x^2+x+a+1,g(x)=x-3`
Áp dụng định lí Bezout có :
`f(3)=2.3^2+3+a+1=22+a`
Để `f(x)` chia hết cho `g(x)`
`=>22+a=0`
`=>a=-22`
Vậy `a=-22` để `2x^2+x+a+1` chia hết cho `x-3`
