phân tích đa thức thành nhân tử:
a) \(xz-yz-x^2+2xy-y^2\)
b) \(x^2+8x+15\)
c) \(x^2-x-12\)
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Những câu hỏi liên quan
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 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
\(x^3-4x^2+4x-xz^2=x\left(x^2-4x+4-z^2\right)\)
\(=x\left[\left(x-2\right)^2-z^2\right]=x\left(x-2-z\right)\left(x-2+z\right)\)
\(x^2-2xy+y^2-z^2+10z-25\)
\(=\left(x-y\right)^2-\left(z-5\right)^2\)
\(=\left(x-y+z-5\right)\left(x-y-z+5\right)\)
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!
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) \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
NT
3
NC
8 tháng 8 2018
1) \(x^2-2xy+y^2-xz+yz\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)-\left(xz-yz\right)\)
\(\Leftrightarrow\left(x-y\right)^2-z\left(x-y\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x-y-z\right)\)
2)\(x^2-y^2-x+y\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+1\right)\)
8 tháng 8 2018
\(a,x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
\(b,x^2-y^2-x+y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
2 tháng 10 2016
\(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
a)xz-yz -x2 +2xy-y2=(xz-yz)-(x2-2xy+y2)=z(x-y)-(x-y)2=(x-y)(z-x+y)
b) x2+8x+15= (x2+3x)+(5x+15)=x(x+3)+5(x+3)=(x+3)(x+5)
c) x2-x-12=(x2-4x)+(3x-12)=x(x-4)+3(x-4)=(x-4)(x+3)
a) xz - yz - x2 + 2xy - y2
= (xz - yz) - (x2 - 2xy + y2)
= z (x - y) - (x - y)2
= z (x - y) - (x - y) (x - y)
= [z - (x - y)] (x - y)
= (z - x + y) (x - y)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= (x2 + 3x) + (5x + 15)
= x (x + 3) + 5 (x + 3)
= (x + 5) (x + 3)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= (x2 - 4x) + (3x - 12)
= x (x - 4) + 3 (x - 4)
= (x + 3) (x - 4)
#Học tốt!!!
~NTTH~