Phân tích đa thức thành nhân tử:
a) 3 x 2 - 3 x y – 5 x + 5 y
b) x 2 + 4 x – 45
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CM
Cao Minh Tâm
3 tháng 5 2018
Đúng(0)
Những câu hỏi liên quan
c) 2x2(x +1) + 4x(x +1); d) 2 x(y - 1) - 2
7 tháng 10 2021
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
KS
7 tháng 10 2021
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
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: \(x^4+4x^2-5\)
\(=x^4+4x^2+4-9\)
\(=\left(x^2+2\right)^2-3^2\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
26 tháng 8 2021
`b)x^3+y^3+z^3-3xyz`
`=x^3+3xy(x+y)+z^3-3xy(x+y)-3xyz`
`=(x+y)^3+z^3-3xy(x+y+z)`
`=(x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y)`
`=(x+y+z)(x^2+2xy+y^2-zx-yz-3xy+z^2)`
`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`
22 tháng 7 2023
`a, P = 2x(3 - x^2)`
`b, Q = 5x^2(x-3y)`
`c, R = xy(3x^2y^2 - 6y^2z + 1)`
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(P=6x-2x^3\)
\(P=2x\left(3+x^2\right)\)
b) \(Q=5x^3-15x^2y\)
\(Q=5x^2\left(x-3y\right)\)
c) \(R=3x^3y^3-6xy^3z+xy\)
\(R=xy\left(3x^2y^2-6y^2z+1\right)\)
22 tháng 7 2023
`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`
`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`
`c, xy^2 + x^2y + 1/4y^3`
`= y(xy + x^2 + 1/4y^2)`
`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`
`= (x+1+y)(x+1-y)`
30 tháng 8 2021
a: Ta có: \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)
\(=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\)
\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)
\(=\left(x-1\right)\left(2x^2-9x+6\right)\)
b: Ta có: \(x\left(y-x\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)\)
\(=-x\left(x-y\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)\)
\(=\left(x-y\right)\left[-x\left(x-y\right)^2-y\left(x-y\right)+xy\right]\)
\(=\left(x-y\right)\left[-x^3+2x^2y-xy^2-xy+y^2+xy\right]\)
\(=\left(x-y\right)\left(-x^3+2x^2y-xy^2+y^2\right)\)
30 tháng 8 2021
a) \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)
b) \(x\left(y-x\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)=\left(x-y\right)\left[-x\left(x-y\right)^2-y\left(x-y\right)+xy\right]=\left(x-y\right)\left(-x^3+2x^2y-xy^2-xy+y^2+xy\right)=\left(x-y\right)\left(-x^3+y^2+2x^2y-xy^2\right)\)
c) \(xy\left(x+y\right)-2x-2y=xy\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(xy-2\right)\)
d) \(x\left(x+y\right)^2-y\left(x+y\right)^2+y^2\left(x-y\right)=\left(x+y\right)^2\left(x-y\right)+y^2\left(x-y\right)=\left(x-y\right)\left(x^2+2xy+y^2+y^2\right)=\left(x-y\right)\left(x^2+2y^2+2xy\right)\)
29 tháng 12 2020
Câu 1:
Phần a đề sai nên mk sửa lại:
a, x2 + 5x - 14 = x2 - 2x + 7x - 14 = x(x - 2) + 7(x - 2) = (x - 2)(x + 7)
b, xz + yz - 5(x + y) = z(x + y) - 5(x + y) = (x + y)(z - 5)
Câu 2:
x2 - 4x = -4
\(\Leftrightarrow\) x2 - 4x + 4 = 0
\(\Leftrightarrow\) (x - 2)2 = 0
\(\Leftrightarrow\) x - 2 = 0
\(\Leftrightarrow\) x = 2
Vậy x = 2
Chúc bn học tốt!
AD
23 tháng 7 2023
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
2
23 tháng 5 2022
`a)(x+2)^2+2(x^2-4)+(x-2)^2`
`=(x+2)^2+2(x-2)(x+2)+(x-2)^2`
`=(x+2+x-2)^2=(2x)^2=4x^2`
`b)x^2-x+1/4`
`=x^2-2.x .1/2+1/4=(x-1/2)^2`
`c)(x+y)^3-(x-y)^3`
`=(x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]`
`=2y(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2)`
`=2y(3x^2+y^2)`
23 tháng 5 2022
a) \(\left(x+2\right)^2+2\left(x^2-4\right)+\left(x-2\right)^2\)
\(=\left(x+2\right)^2+2\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(x+2+x-2\right)^2=\left(2x\right)^2=4x^2\)
b) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
c) \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)