a)  \(x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)

b)  \(x^3-3x^2y-x+3y=x^2\left(x-3y\right)-\left(x-3y\right)=\left(x-3y\right)\left(x-1\right)\left(x+1\right)\)

c)  \(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)

help me

\(2x^2+x-4xy-2y+2x+1=\left(2x^2+x\right)-\left(4xy+2y\right)+\left(2x+1\right)=x\left(2x+1\right)-2y\left(2x+1\right)+\left(2x+1\right)=\left(x-2y+1\right)\left(2x+1\right)\)

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

a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2

1)

a) \(-7x\left(3x-2\right)\)

\(=-21x^2+14x\)

b) \(87^2+26.87+13^2\)

\(=87^2+2.87.13+13^2\)

\(=\left(87+13\right)^2\)

\(=100^2\)

\(=10000\)

2)

a) \(x^2-25\)

\(=x^2-5^2\)

\(=\left(x-5\right)\left(x+5\right)\)

b) \(3x\left(x+5\right)-2x-10=0\)

\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)

\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy..........

3)

a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)

Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)

b)

Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)

Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)

Bài `1:`

`a)3x^3+6x^2=3x^2(x+2)`

`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`

Bài `2:`

`a)(2x-1)^2-25=0`

`<=>(2x-1-5)(2x-1+5)=0`

`<=>(2x-6)(2x+4)=0`

`<=>[(x=3),(x=-2):}`

`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`

`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`

`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`

`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`

C1

a) -7x(3x-2)=-21x^2+14x

b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2

C2

a) (x-5)(x+5)

b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0

\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)

Vậy S={-5;2/3}

C3:

a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3

b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)

\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)

Bài 1:

a: \(3x-6y=3\cdot x-3\cdot2y=3\left(x-2y\right)\)

b: \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)

\(=7xy\left(2x-3y+4xy\right)\)

c: \(10x\left(x-y\right)-8y\cdot\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=\left(2\cdot5x+2\cdot4y\right)\left(x-y\right)\)

\(=2\left(5x+4y\right)\left(x-y\right)\)

bài 2:

a: Đề thiếu vế phải rồi bạn

b: \(x^3-13x=0\)

=>\(x\left(x^2-13\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{13}\end{matrix}\right.\)

Bài 1:

a, $3x-6y$

$=3(x-2y)$

b, $14x^2y-21xy^2+28x^2y^2$

$=7xy(2x-3y+4xy)$

c, $10x(x-y)-8y(y-x)$

$=10x(x-y)-8y[-(x-y)]$

$=10x(x-y)+8y(x-y)$

$=(x-y)(10x+8y)$

$=2(x-y)(5x+4y)$

Bài 2:

a, Đề thiếu rồi bạn nhé.

b, \(x^3-13x=0\)

\(\Rightarrow x\left(x^2-13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\)

kết quả thôi nha

umk nhanh nha bạn

2:

a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)

\(=x\left(2x-y\right)+z\left(x-2y\right)=\left(x-2y\right)\left(x+z\right)\)

b: \(=\left(x^2-4y^2\right)-\left(x-2y\right)\)

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

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

c: \(=\left(y^2+10y+25\right)-9z^2\)

\(=\left(y+5\right)^2-\left(3z\right)^2\)

\(=\left(y+5+3z\right)\left(y+5-3z\right)\)

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

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

\(=\left(x+2y\right)\left(x^2+4xy+4y^2-x+2y\right)\)

1:

a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)

b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)

\(=2y\left(5y-6\right)+4\left(5y-6\right)\)

\(=2\left(5y-6\right)\left(y+2\right)\)

c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)

\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)

\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)

d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)

\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)

\(=2y\left(x+y\right)\left(3x-3y-4x-4y\right)\)

\(=2y\left(x+y\right)\left(-x-7y\right)\)

Bài 1

a) x(3 - 4x) + 5(3 - 4x)

= (3 - 4x)(x + 5)

b) 2y(5y - 6) - 4(6- 5y)

= 2y(5y - 6) + 4(5y - 6)

= (5y - 6)(2y + 4)

= 2(5y - 6)(y + 2)

c) 27(x - 2)³ - 3x(2 - x)²

= 27(x - 2)³ - 3x(x - 2)²

= 3(x - 2)²[9(x - 2) - x]

= 3(x - 2)²(9x - 18 - x)

= 3(x - 2)²(8x - 18)

= 6(x - 2)²(4x - 9)

d) 6y(x² - y²) - 8y(x + y)²

= 6y(x - y)(x + y) - 8y(x + y)²

= 2y(x + y)[3(x - y) - 4(x + y)]

= 2y(x + y)(3x - 3y - 4x - 4y)

= 2y(x + y)(-x - 7y)

= -2y(x + y)(x + 7y)