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e: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
f: \(x^2-2x+7x-14\)
\(=x\left(x-2\right)+7\left(x-2\right)\)
=(x-2)(x+7)
h: \(5x^2-10xy+5y^2-20\)
\(=5\left(x^2-2xy+y^2-4\right)\)
\(=5\left(x-y-2\right)\left(x-y+2\right)\)
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a: \(3x^4-6x^3+2x^2=x^2\left(3x^2-6x+2\right)\)
b: \(x^3y+12x^2y+36xy=xy\left(x^2+12x+36\right)=xy\left(x+6\right)^2\)
c: \(x^3y-9xy^3=xy\left(x^2-9y^2\right)=xy\left(x-3y\right)\left(x+3y\right)\)
d: \(x^2y^2-2xy^2+y^2=y^2\left(x-1\right)^2\)
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a: =x^3(x-y)+(x-y)
=(x-y)(x^3+1)
=(x-y)(x+1)(x^2-x+1)
b: =(a-1)^2-9b^2
=(a-1-3b)(a-1+3b)
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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, \(\left(x+y\right)\left(x-1\right)\)
\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(A=10x^2+20xy+10y^2-90\)
\(=10\left(x^2+2xy+y^2-9\right)\)
\(=10\left(x+y-3\right)\left(x+y+3\right)\)
b: Ta có: \(B=x^3y-3x^2y-4xy+12y\)
\(=x^2y\left(x-3\right)-4y\left(x-3\right)\)
\(=y\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
c: Ta có: \(C=125x^3-10x^2+2x-1\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)
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1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(2x^3-3x^2=x^2\left(2x-3\right)\)
b. \(3x^4-24x=3x\left(x^3-8\right)=3x\left(x-2\right)\left(x^2+2x+4\right)\)
c. \(x^3y+5x^2y=x^2y\left(x+5\right)\)
d. \(7x^2+14xy=7x\left(x+2y\right)\)
a)2x^3-3x^2=x^2(2x-3)
b)3x^4-24x
=3x(x^3-8x)
=3x(x-2)(x^2+2x+4)
c)x^3y+5x^2y
=x^2y(x+5)
d)7x^2+14xy
=7x(x+2y)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
A=x^3y(x^4+x^2+1)
Hết
x7y+x5y+x3y=x3y.(x4+x2+1)
=x3y.(x4+2x2+1-x2)
=x3y.[(x2+1)2-x2]
=x3y.(x2-x+1)(x2+x+1)