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)

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

\(=xy\left(x^2-2xy+5\right)\)

a) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)

b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(3x+5\right)\left(x-y\right)\)

Vô đây xem: bài 1:phân tích đa thức thành nhân tửa)7x^3y-14x^2y+7xy^3b)3x^2-3xy-5x+5yc)x^2+7x+12giúp mình với  - Hoc24

\(1,\\ a,=2x^2+2x\\ b,=x^2+4x+3-4=x^2+4x-1\\ c,=x^2+4x+4+3x-5=x^2+7x-1\\ 2,\\ a,=3\left(x+y\right)\\ b,=\left(x-3\right)^2\\ c,=7\left(x+y\right)\\ 3,\\ \Leftrightarrow\left(x-1\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

a) 2x²+2x

a: \(=\dfrac{2}{5}\left(xy-x-y^2+1\right)\)

\(=\dfrac{2}{5}\left[x\left(y-1\right)-\left(y-1\right)\left(y+1\right)\right]\)

\(=\dfrac{2}{5}\left(y-1\right)\left(x-y-1\right)\)

b: \(=x\left(x^2+2xy+y^2-9\right)\)

\(=x\left(x+y-3\right)\left(x+y+3\right)\)

=25(y−1)(x−y−1)=25(y−1)(x−y−1)

b) $=x\left({}^{}+2xy+{}^{}-9\right)$

$=x\left(x+y-3\right)\left(x+y+3\right)$

`a)7x^3y^2+14x^2y^3+7xy^4`

`=7xy^2(x^2+2xy+y^2)`

`=7xy^2(x+y)^2`

______________________________________________

`b)x^2-xy+5x-5y`

`=x(x-y)+5(x-y)`

`=(x-y)(x+5)`

______________________________________________

`c)3x^2-6xy-12+3y^2`

`=3(x^2-2xy-4+y^2)`

`=3[(x-y)^2-4]`

`=3(x-y-2)(x-y+2)`

a)7x³y²+14x²y³+7xy⁴

=7xy²(x²+2xy+y²)

=7xy²(x+y)²

b)x²-xy + 5x - 5y

=x(x-y) + 5(x-y)

=(x-y) (x+5)

p

\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)

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) $9x^2-6x-y^2+2y$

$=(9x^2-y^2)-(6x-2y)$

$=[(3x)^2-y^2]-2(3x-y)$

$=(3x-y)(3x+y)-2(3x-y)$

$=(3x-y)(3x+y-2)$

$\text{#}Toru$

bạn ấn ở chỗ x² cho rõ hơn nhé

\(5x^2+10xy=5x\left(x+2y\right)\)

\(x^2+xy-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\)

\(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)

\(x^2-7x+6=x^2-x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)