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phân tích đa thức thành nhân tử
a/4x-3x-1
b/x^7+x^5+1
c/x^3-x^2-4
d/3x^3-7x^2+17x-5
e/x^2+2xy+y^2-x-y-12
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
\(x^2+2xy+y^2-x-y-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)
\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)=\left(x+y+3\right)\left(x+y-4\right)\) \(P=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (nhóm 2 cái đầu với cuối lại với nhau, 2 cái giữa vào 1 nhóm)
Đặt \(x^2+7x+11=a\)
Ta có: \(P=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-25=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
d, \(4x^4-32x^2+1\)
\(=4x^4+4x^2+1-36x^2\)
\(=\left(2x+1\right)^2-\left(6x\right)^2=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5x^2 + 6xy + y^2
=5x2+5xy+xy+y2
=5x.(x+y)+y.(x+y)
=(x+y)(5x+y)
b) x^2 + 2xy - 15y^2.
=x2-3xy+5xy-15y2
=x.(x-3y)+5y.(x-3y)
=(x-3y)(x+5y)
c) (x-y)^2 + 4(x-y) - 12
=(x-y)2+4(x-y)+4-16
=(x-y+2)2-16
=(x-y+2-4)(x-y+2+4)
=(x-y-2)(x-y+6)
d) x^3 - 2x - 4.
=x3+2x2+2x-2x2-4x-4
=x.(x2+2x+2)-2.(x2+2x+2)
=(x2+2x+2)(x-2)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) =(x2+x-1)2 -16=(x2+x-5)(x2+x+3)
b) Đặt a=x2+3x
=(a+1)(a+2)-6=a2+3a-4=(a-1)(a+4)=(x2+3x-1)(x2+3x+4)
c) =(x+y)2-(x+y)-12=(x+y)(x+y+1)-12
Đặt a=x+y
=a(a+1)-12=a2+a-12=(a-3)(a+4)=(x+y-3)(x+y+4)
d) Đặt a=x2+x
=(a+1)(a+2)-12=a2+3a-10=(a-2)(a+5)=(x2+x-2)(x2+x+5)
![](https://rs.olm.vn/images/avt/0.png?1311)
2/a/ \(x\left(x+1\right)^2\)
b/ \(\left(y+x\right)\left(y-1\right)\)
c/ \(\left(x+1\right)\left(3x-10\right)\)
d/ \(-3\left(2z+x-y\right)\left(2z+y-x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(x^2+2xy+y^2-x-y-12\)
= \(\left(x+y\right)^2-\left(x+y\right)-12\)
Đặt \(x+y=z\) (đặt ẩn phụ)
\(\Rightarrow z^2-z-12\)
\(=z^2+3z-4z-12\)
\(=z\left(z+3\right)-4\left(z+3\right)\)
\(=\left(z+3\right)\left(z-4\right)\)
Khi đó: \(\left(x+y+3\right)\left(x+y-4\right)\)
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