1 phân tích đa thức thành nhân tử
a,x^2+2x(y+1)+1+2y+y^2
b, x^2+9x+8
c,x+x-42
d, x^2+9x+8
2 cmr:
A=(2020^5-2020^3) chia hết 2019*2020*2021
1 phân tích đa thức thành nhân tử
a,x^2+2x(y+1)+1+2y+y^2
b, x^2+9x+8
c,x+x-42
d, x^2+9x+8
2 cmr:
A=(2020^5-2020^3) chia hết 2019*2020*2021
Bái 1:
a,x²+2x(y+1)+1+2y+y²=x²+2x(y+1)+(y+1)²
=(x+y+1)²
b,x²+9x+8=x²+x+8x+8
=x(x+1)+8(x+1)
=(x+1)(x+8)
c,x+x-42=2x-42=2(x-21)
d,x²+9x+8=x²+x+8x+8
=x(x+1)+8(x+1)
=(x+1)(x+8)
Bài 2:
A=(2020^5-2020³)
A=2020³(2020²-1)
A=2020³(2020-1)(2020+1)
A=2020³.2019.2021
Thấy (2020³.2019.2021):(2020.2019.2021)=2020²
Vậy A chia hết cho 2019.2020.2021(đpcm)
`1`
`a, x^2 +2x(y+1)+1+2y+y^2`
`=x^2 +2x(y+1)+(y^2 +2y+1)`
`=x^2 +2x(y+1)+(y+1)^2`
`=(x+y+1)^2`
`b, x^2 +9x+8`
`=x^2 +x+8x+8`
`=(x^2 +x)+(8x+8)`
`=x(x+1)+8(x+1)`
`=(x+1)(x+8)`
`c, x+x-42`
`=2x-42`
`=2(x-21)`
`d, x^2 +9x+8`
`=x^2 +x+8x+8`
`=(x^2 +x)+(8x+8)`
`=x(x+1)+8(x+1)`
`=(x+1)(x+8)`
`2`
`A=2020^5 -2020^3`
`A=2020^3 (2020^2 -1)`
`A=2020^3 (2020-1)(2020+1)`
`A=2020^3 .2019.2021`
Ta có: `2020^3 \vdots2020 , 2019\vdots2019,2021\vdots2021`
`=> 2020^3 .2019.2021\vdots2019.2020.2021`
`=> A\vdots2019.2020.2021`
chúc học tốt