Phân tích đa thức thành nhân tử
\(x^3+3x^2y-9xy^2+5y^2\)
\(x^8y^8+x^4y^4+1\)
K
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29 tháng 8 2018
mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)
KN
9 tháng 9 2019
a) \(x^3y^3+x^2y^2+4\)
\(=x^3y^3-x^2y^2+2x^2y^2-2xy+2xy+4\)
\(=\left(x^3y^3-x^2y^2+2xy\right)+\left(2x^2y^2-2xy+4\right)\)
\(=xy\left(x^2y^2-xy+2\right)+2\left(x^2y^2-xy+2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
b) \(x^3+3x^2y-9xy^2+5y^3\)
\(=x^3+5x^2y-2x^2y-10xy^2+xy^2+5y^3\)
\(=\left(5y^3-10xy^2+5x^2y\right)+\left(xy^2-2x^2y+x^3\right)\)
\(=5y\left(y^2-2xy+x^2\right)+x\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y-x\right)^2\)
LT
1
PN
12 tháng 10 2017
x3 + 3x2y - 9xy2 + 5y3
= ( x3 - 3x2y + 3xy2 - y3 ) + ( 6y3 - 12xy2 + 6 x2y )
= ( x - y )3 + 6y ( x - y )2
= ( x - y )2 ( x + 5y )
9 tháng 10 2021
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
29 tháng 9 2015
a,3x3y3-15x2y2=3x2y2(xy-5)
b,2x(x-5y)+8y(5y-x)=2x(x-5y)-8y(x-5y)=(x-5y).(2x-8y)
c,(3x-1)2-16=(3x-1)2-42=(3x-1+4)(3x-1-4)=(3x+3)(3x-5)
d,x3-3x2+3x-1=x3-1-(3x2+3x)=x3-1-3x(x+1)=(x3-1-3x)(x+1)
e,125x3+1=(5x)3+13=(5x+1)(25x2-5x.1+12)
f,x3+6x2y+12xy2+8y3=x3+3.x2.2y+3.x.(2y)2+(2y)3=(x+2y)3
6 tháng 8 2021
a, \(x-2y+x^2-4y^2=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(1+x+2y\right)\)
b, \(x^2-4x^2y^2+y^2+2xy=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)
6 tháng 8 2021
a) Ta có: \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(1+x+2y\right)\)
b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
\(x^8y^8+x^4y^4+1=\left[\left(x^4y^4\right)^2+2x^4y^4+1\right]-x^4y^4=\left(x^4y^4+1\right)^2-\left(x^2y^2\right)^2\)
\(=\left(x^4y^4+1-x^2y^2\right)\left(x^4y^4+1+x^2y^2\right)\)
\(=\left(x^4y^4+1-x^2y^2\right)\left[\left(x^2y^2\right)^2+2x^2y^2+1-x^2y^2\right]\)
\(=\left(x^4y^4+1-x^2y^2\right)\left[\left(x^2y^2+1\right)^2-\left(xy\right)^2\right]\)
\(=\left(x^4y^4+1-x^2y^2\right)\left(x^2y^2+1-xy\right)\left(x^2y^2+1+xy\right)\)
Phân tích đa thức thành nhân tử
x3+3x2y−9xy2+5y2
x8y8+x4y4+1