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a: \(x^4+2x^3+x^2=x^2\left(x+1\right)^2\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
e) Ta có: \(a^3-a^2-a+1\)
\(=a^2\left(a-1\right)-\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2-1\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)\)
f) Ta có: \(x^3-2xy-x^2y+2y^2\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
a) \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2+2ab\right)\left(a^2+b^2-2ab\right)=\left(a+b\right)^2.\left(a-b\right)^2\)
b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)
d) Đề sai ko ???
e) \(a^3-a^2-a+1=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)=\left(a-1\right)^2\left(a+1\right)\)
f) \(x^3-2xy-x^2y+2y^2=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =x^2(x^2+2x+1)
=x^2(x+1)^2
b: =x^3+3x^2y+3xy^2+y^3-x-y
=(x+y)^3-(x+y)
=(x+y)[(x+y)^2-1]
=(x+y)(x+y-1)(x+y+1)
c: =5(x^2-2xy+y^2-4z^2)
=5(x-y-2z)(x-y+2z)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
Bài 3:
a: \(2x\left(x-3\right)-x+3=0\)
=>\(2x\left(x-3\right)-\left(x-3\right)=0\)
=>(x-3)(2x-1)=0
=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)
=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)
=>\(5x^2-x-2=5x^2\)
=>-x-2=0
=>-x=2
=>x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)