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1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) $4x^2+4x+1$
$=(2x)^2+2\cdot2x\cdot1+1^2$
$=(2x+1)^2$
b) $x^2+6x-y^2+9$
$=(x^2+6x+9)-y^2$
$=(x^2+2\cdot x\cdot3+3^2)-y^2$
$=(x+3)^2-y^2$
$=(x+3-y)(x+3+y)$
$\text{#}Toru$
a: \(4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)
\(=\left(2x+1\right)^2\)
b: \(x^2+6x-y^2+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3+y\right)\left(x+3-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=3xy\left(x-2y\right)\\ b,=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x+y+3\right)\left(x-y\right)\\ c,=x\left[\left(x+2\right)^2-y^2\right]=x\left(x+y+2\right)\left(x-y+2\right)\\ d,\Leftrightarrow x\left(x^2-4\right)=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
e) \(8\left(x+3y\right)-16x\left(x+3y\right)=\left(x+3y\right)\left(8-16x\right)=8\left(x+3y\right)\left(1-2x\right)\)
f) \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=\left(x+1\right)\left(4x^2+2x^2\right)=6x^2\left(x+1\right)\)
g) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)