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a, \(x-2y+x^2-4y^2=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(1+x+2y\right)\)
b, \(x^2-4x^2y^2+y^2+2xy=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)
a) Ta có: \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(1+x+2y\right)\)
b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ a,=2x^2+2x\\ b,=x^2+4x+3-4=x^2+4x-1\\ c,=x^2+4x+4+3x-5=x^2+7x-1\\ 2,\\ a,=3\left(x+y\right)\\ b,=\left(x-3\right)^2\\ c,=7\left(x+y\right)\\ 3,\\ \Leftrightarrow\left(x-1\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
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a) $x^3-3x^2y+4x-12y$
$=(x^3-3x^2y)+(4x-12y)$
$=x^2(x-3y)+4(x-3y)$
$=(x-3y)(x^2+4)$
b) $4x^2-y^2+4y-4$
$=4x^2-(y^2-4y+4)$
$=(2x)^2-(y^2-2\cdot y\cdot2+2^2)$
$=(2x)^2-(y-2)^2$
$=[2x-(y-2)][2x+(y-2)]$
$=(2x-y+2)(2x+y-2)$
c) $9x^2-6x-y^2+2y$
$=(9x^2-y^2)-(6x-2y)$
$=[(3x)^2-y^2]-2(3x-y)$
$=(3x-y)(3x+y)-2(3x-y)$
$=(3x-y)(3x+y-2)$
$\text{#}Toru$
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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)
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2:
a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)
b: \(2\left(x-1\right)+x^2-x\)
\(=2\left(x-1\right)+x\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
c: \(3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)
3:
a: \(2x\left(x-1\right)-2x^2=4\)
=>\(2x^2-2x-2x^2=4\)
=>-2x=4
=>x=-2
b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)
=>\(x^2-3x-\left(x^2+x-2\right)=5\)
=>\(x^2-3x-x^2-x+2=5\)
=>-4x=3
=>x=-3/4
c: \(4x^2-25+\left(2x+5\right)^2=0\)
=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)
=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)
=>4x(2x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)
a) 2x2-2y2-6x-6y = ( 2x2-2y2)- ( 6x+6y)
= 2(x2-y2)- 6(x+y)
2( x+y )- 6( x+y )
..........
b) x^3 +3x^2 - 3x -1= (x^3-1) +(3x^2-3x)
= (x3-1)+ 3x( x-y)
.......
Những dòng mình .... là đến đấy đơn giản rồi, đặt nhân tử chung là đc
=
#Vũ Khánh Linh câu a/ dòg thứ 3 hình như sai sai.... @@