Phân tích thành nhân tử:
a, x^2-x-y^2-y
b, x^2-2xy+y^2-z^2
K
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HP
25 tháng 10 2021
a. 16xy2 - 12xy + 24x2y
= 4xy(4y - 3 + 6x)
c. 16 - x2 + 2xy - y2
= 42 - (x2 - 2xy + y2)
= 42 - (x - y)2
= (4 - x + y)(4 + x - y)
25 tháng 10 2021
b: \(x^3-x^2-x+1=\left(x-1\right)^2\left(x+1\right)\)
d: \(x^2-x-20=\left(x-5\right)\left(x+4\right)\)
AH
2
12 tháng 12 2021
\(a,=5x\left(x-1\right)+y\left(x-1\right)=\left(5x+y\right)\left(x-1\right)\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
6 tháng 1 2022
a)5x2-5x+xy-y=(5x2-5x)+(xy-y)=5x(x-1)+y(x-1)=(x-1)(5x+y)
b)x2-2xy+y2-9=(x2-2xy+y2)-9=(x-y)2-32=(x-y-3)(x-y+3)
22 tháng 10 2021
\(a,=x\left(x^2-4x+4-z^2\right)=x\left[\left(x-2\right)^2-z^2\right]=x\left(x-z-2\right)\left(x+z-2\right)\\ b,=\left(x-y\right)^2-\left(z-5\right)^2=\left(x-y-z+5\right)\left(x-y+z-5\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 10 2021
\(x^3-4x^2+4x-xz^2=x\left(x^2-4x+4-z^2\right)\)
\(=x\left[\left(x-2\right)^2-z^2\right]=x\left(x-2-z\right)\left(x-2+z\right)\)
\(x^2-2xy+y^2-z^2+10z-25\)
\(=\left(x-y\right)^2-\left(z-5\right)^2\)
\(=\left(x-y+z-5\right)\left(x-y-z+5\right)\)
10 tháng 12 2021
\(a,=xy\left(x+2y+1\right)\\ b,=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\\ c,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ d,=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=\left(x-2\right)\left(x+2+x-2\right)=2x\left(x-2\right)\\ e,=\left(x+1\right)^2-y^2=\left(x+y+1\right)\left(x-y+1\right)\\ g,=\left(x+9-6x\right)\left(x+9+6x\right)=\left(9-5x\right)\left(7x+9\right)\\ h,=\left(x-y\right)^2-\left(z-t\right)^2=\left(x-y-z+t\right)\left(x-y+z-t\right)\\ i,=\left(x-1\right)^3-y^3=\left(x-y-1\right)\left(x^2-2x+1+xy+y+y^2\right)\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(x^2-xy+x-y\)
\(=\left(x^2+x\right)-\left(xy+y\right)\)
\(=x\left(x+1\right)-y\left(x+1\right)\)
\(=\left(x+1\right)\left(x-y\right)\)
b) \(x^2+2xy-4x-8y\)
\(=x\left(x+2y\right)-4\left(x+2y\right)\)
\(\left(x-4\right)\left(x+2y\right)\)
c) \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
G
1
6 tháng 1 2022
a:=2(x-2)+y(x-2)
=(x-2)(y+2)
b: \(=\left(x+y\right)^2-4\)
=(x+y+2)(x+y-2)
SO
2
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 7 2021
a.
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)
\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)
b.
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c.
\(=x^4-1+4x^2-4\)
\(=\left(x^2-1\right)\left(x^2+1\right)+4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
16 tháng 7 2021
a) Ta có: \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
b) Ta có: \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+4x^2+4-9\)
\(=\left(x^2+2\right)^2-3^2\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
22 tháng 7 2023
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
a) \(x^2-x-y^2-y\)
\(=\left(x-y\right).1\)
b) \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-x^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
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