phân tích thành nhân tử
1) 2x2(x+1) + 4(x+1)
2) -3x - 6xy - 9xz
3) 2x2y - 4xy2 + 6xy
4) 4x3y2 - 8x3y2 + 2x4y
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Những câu hỏi liên quan
10 tháng 3 2022
1, \(xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(x+y\right)\)
2, \(5x\left(3y+4x-6\right)\)
3, \(3x\left(2-y\right)\)
4, \(x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)
5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)
6, \(2xy\left(x+2y-5x^2y\right)\)
7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
11, \(\left(x+y\right)\left(x-1\right)\)
VD
10 tháng 3 2022
\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)
26 tháng 10 2021
a: \(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(2x+3\right)\left(x^2+1\right)\)
b: \(=\left(x-4\right)\left(x+3\right)\)
e: =(x+3)(x-2)
26 tháng 10 2021
a) \(=x^2\left(2x+3\right)+\left(2x+3\right)=\left(2x+3\right)\left(x^2+1\right)\)
b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)
c) \(=\left(2x\right)^2-\left(x^2+1\right)^2=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=4xy\left(y-3x+2\right)\)
e) \(=x\left(x-2\right)+3\left(x-2\right)=\left(x-2\right)\left(x+3\right)\)
f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)
g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)
h) \(=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)
i) \(=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)^2\left(x+3\right)\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
4 tháng 11 2023
Bài 1:
\(a,2x^2y\left(2x^2y^2-xy^2\right)\\ =2x^2x^2y^2y-2x^2x.y^2.y=2x^4y^3-2x^3y^3\\ b,\left(x-1\right)\left(2x+3\right)\\ =x.2x+x.3-1.2x-1.3=2x^2+3x-2x-3\\ =2x^2+x-3\\ c,\left(20x^3y^4+10x^2y^3-5xy\right):5xy\\ =20x^3y^4:5xy+10x^2y^3:5xy-5xy:5xy\\ =\left(20:5\right).\left(x^3:x\right).\left(y^4:y\right)+\left(10:5\right).\left(x^2:x\right).\left(y^3:y\right)-\left(5:5\right).\left(x:x\right).\left(y:y\right)\\ =4x^2y^3+2xy^2-1\\ d,\left(y-3x\right)^2-\left(y^2-6xy\right)\\ =\left[y^2-2.y.3x+\left(3x\right)^2\right]-\left(y^2-6xy\right)\\ =y^2-6xy+9x^2-y^2+6xy =9x^2\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
4 tháng 11 2023
Bài 2:
\(a,4xy+4xz=4x\left(y+z\right)\\ b,x^2-y^2+9-6x\\ =\left(x^2-6x+9\right)-y^2\\ =\left(x-3\right)^2-y^2\\ =\left(x-3-y\right)\left(x-3+y\right)\)
Bài 3:
\(a,\dfrac{3xy}{y+z}+\dfrac{3xz}{y+z}\\=\dfrac{3xy+3xz}{y+z}\\ =\dfrac{3x\left(y+z\right)}{\left(y+z\right)}=3x\left(Với:y\ne-z\right)\\ b,\dfrac{x}{x+2}-\dfrac{x}{x-2}\\ =\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x^2-2x}{\left(x+2\right)\left(x-2\right)}=0\)
16 tháng 1 2023
1/2x^3y(2x^4y^3-4xy-6)
=1/2x^3y*2x^4y^3-1/2x^3y*4xy-1/2x^3y*6
=x^7y^4-2x^4y^2-3x^3y
16 tháng 1 2023
1/2x^3y(2x^4y^3-4xy-6)
=1/2x^3y*2x^4y^3-1/2x^3y*4xy-1/2x^3y*6
=x^7y^4-2x^4y^2-3x^3y
16 tháng 1 2023
1: \(=-3x^3-21x^2+x\)
2: \(=-15x^4y^7+10x^5y^6+5x^3y^5\)
3: \(=x^7y^4-2x^4y^2-3x^3y\)
5: \(=15x-6x^2\)
6: \(=4x^3-8x^2+10x\)
7: \(=-8x^5y^3+16x^7y^2-12x^3y^4\)
8: \(=x^7y^4-2x^4y^2-3x^3y\)
12 tháng 7 2021
a) Ta có: \(a^3y^3+125\)
\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)
b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy-6xy+y^2\right)\)
\(=\left(2x-y\right)^3\)
NN
4
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 9 2023
b, (\(x^2\) - \(xy\) ) + (\(x-y\))
= (\(x-y\)).\(x\) + (\(x-y\))
= (\(x-y\)).(\(x\) + 1)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 9 2023
c, \(x^2\) - 2\(x\) - 15
= (\(x^2\) - 2\(x\) + 1) - 16
= (\(x\) - 1)2 - 42
= (\(x-1-4\)).(\(x-1+4\))
= (\(x-5\)).(\(x+3\))
\(2x^2\left(x+1\right)+4\left(x+1\right)=2\left(x+1\right)\left(x^2+2\right)\)
\(-3x-6xy-9xz=-3x\left(1+2y+3z\right)\)
\(2x^2y-4xy^2+6xy=2xy\left(x-2y+3\right)\)
\(4x^3y^2-8x^3y^2+2x^4y=-4x^3y^2+2x^4y=2x^3y\left(x-2y\right)\)
1) \(2x^2\left(x+1\right)+4\left(x+1\right)=2\left(x+1\right)\left(x^2+2\right)\)
2) \(-3x-6xy-9xz=-3x\left(1+2y+3z\right)\)
3) \(2x^2y-4xy^2+6xy=2xy\left(x-2y+3\right)\)
4) \(4x^3y^2-8x^3y^2+2x^4y=-4x^3y^2+2x^4y=-2x^3y\left(2y-x\right)\)