Phân tích các đa thức sau thành nhân tử:
a) xy+y-2x-2
b) xy+1+x+y
c) x2+xy-x-y+xz-z
K
Khách
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Những câu hỏi liên quan
H
19 tháng 12 2023
Bài `1`
\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)
Bài `3`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
AH
Akai Haruma
Giáo viên
6 tháng 1
Lời giải:
a. $5x^2-10xy=5x(x-2y)$
b. $3x(x-y)-6(x-y)=(x-y)(3x-6)=3(x-y)(x-2)$
c. $2x(x-y)-4y(y-x)=2x(x-y)+4y(x-y)=(x-y)(2x+4y)=2(x-y)(x+2y)$
d. $9x^2-9y^2=9(x^2-y^2)=9(x-y)(x+y)$
e. $x^2-xy-x+y=(x^2-xy)-(x-y)=x(x-y)-(x-y)=(x-y)(x-1)$
f. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$
DP
1
19 tháng 10 2021
\(b,=x^2-xy=x\left(x-y\right)\\ c,=y\left(x+1\right)+z\left(x+1\right)=\left(y+z\right)\left(x+1\right)\\ d,=x^2+ax+bx+ab\\ =x\left(x+a\right)+b\left(x+a\right)=\left(x+b\right)\left(x+a\right)\)
1 tháng 9 2019
a) \(xy+y-2x-2\)
\(=y\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(y-2\right)\)
b) \(xy+1+x+y\)
\(=y\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(y+1\right)\)
c) \(x^2+xy-x-y+xz-z\)
\(=\left(x^2-x\right)+\left(xy-y\right)+\left(xz-z\right)\)
\(=x\left(x-1\right)+y\left(x-1\right)+z\left(x-1\right)\)
\(=\left(x-1\right)\left(x+y+z\right)\)
22 tháng 10 2021
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
NT
0
PA
14 tháng 9 2016
x2 - x - y2 - y
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
***
9x2 + y2 - 16z2 + 6xy
= (3x + y)2 - (4z)2
= (3x + y - 4z)(3x + y + 4z)
***
a3 - a2x - ay + xy
= a2(a - x) - y(a - x)
= (a - x)(a2 - y)
***
2x2 - 8y2 + 3x + 6y
= 2(x2 - 4y2) + 3(x + 2y)
= 2(x - 2y)(x + 2y) + 3(x + 2y)
= (x + 2y)(2x - 4y + 3)
***
xy(x + y) + yz(y + z) + xz(x + z) + 2xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + z)
= y(x + y + z)(x + z) + xz(x + z)
= (x + z)(xy + y2 + yz + xz)
= (x + z)[y(x + y) + z(x + y)]
= (x + z)(x + y)(y + z)
10 tháng 10 2021
h) \(y\left(y-x\right)^3-x\left(x-y\right)^2+xy\left(x-y\right)=y\left(y-x\right)^3-x\left(y-x\right)^2-xy\left(y-x\right)=\left(y-x\right)\left[y\left(y-x\right)^2-x-xy\right]=\left(y-x\right)\left[y\left(y^2-2xy+x^2\right)-x-xy\right]=\left(y-x\right)\left(y^3-2xy^2+x^2y-x-xy\right)\)
i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)
HT
29 tháng 7 2016
\(a,49.\left(y-4\right)^2-9y^2-36y-36=49\left(y-4\right)^2-9\left(y^2+4y+4\right)\)
\(=49\left(y-4\right)^2-9\left(y+4\right)^2=\left(7y-28\right)^2-\left(3y+12\right)^2\)
\(=\left(7y-28+3y+12\right)\left(7y-28-3y-12\right)\)
\(=\left(10y-16\right)\left(4y-40\right)=8\left(5y-8\right)\left(y-10\right)\)
\(b,xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=xyz-xy-yz-xz+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-x-y+1\right)\)
\(=\left(z-1\right)\text{[}x\left(y-1\right)-\left(y-1\right)\text{]}\)
\(=\left(z-1\right)\left(y-1\right)\left(x-1\right)\)
a) \(xy+y-2x-2\)
\(=y\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(y-2\right)\)
b) \(xy+1+x+y\)
\(=y\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(y+1\right)\)
c) \(x\left(x-1\right)+y\left(x-1\right)+z\left(x-1\right)\)
\(=\left(x-1\right)\left(x+y+z\right)\)