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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) \(x^2+xy-x-y+xz-z\)
\(=\left(x^2-x\right)+\left(xy-y\right)+\left(xz-z\right)\)
\(=x\left(x-1\right)+y\left(x-1\right)+z\left(x-1\right)\)
\(=\left(x-1\right)\left(x+y+z\right)\)
NH
7 tháng 10 2019
a, \(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)\(=x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+2xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z\right)\)
\(=x\left(x+y+z\right)\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+z\right)+y\left(x+z\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b, \(2x^2+2y^2-x^2z+z-y^2z-2\)
\(=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)\)
\(=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)\)
\(=\left(2-z\right)\left(x^2+y^2-1\right)\)
15 tháng 7 2016
Câu a:
Cách 1:
\(xy+y-2x-2\)
\(\Leftrightarrow\left(xy+y\right)-\left(2x+2\right)\)
\(\Leftrightarrow y\left(x+1\right)-2\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(y-2\right)\)
Cách 2:
\(xy+y-2x-2\)
\(\Leftrightarrow\left(xy-2x\right)+\left(y-2\right)\)
\(\Leftrightarrow x\left(y-2\right)+\left(y-2\right)\)
\(\Leftrightarrow\left(x+1\right)\left(y-2\right)\)
15 tháng 7 2016
a)xy+y-2x-2
y(x + 1) - 2(x + 1)
<=> (y - 2)(x + 1)
b) x + x + x + 1
<=> 3x + 1
c)x3-3x2+3x-9
<=>(x - 3 ) 3
d)xy+xz+y2+yz
<=> x(y + z) + y(y + z)
<=> (x + y)(y + z)
f)x2+xy+xz-x-y-z
<=> x(x - y -z +1
NA
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)
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`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
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) \(x\left(x-1\right)+y\left(x-1\right)+z\left(x-1\right)\)
\(=\left(x-1\right)\left(x+y+z\right)\)