b) Ta có: \(x^3-x^2y-xy^2+y^3\)

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

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

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

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

Mai cho bn đấy tui dg định off =))

a)\(11x+11y-x^2-xy\)

\(=\left(11x+11y\right)-\left(x^2+xy\right)\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(11-x\right)\left(x+y\right)\)

b)\(x^2-xy-8x+8y\)

\(=\left(x^2-xy\right)-\left(8x-8y\right)\)

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

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

c)\(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

d)\(x^2+2xy+y^2-xz-yz\)

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

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

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

a) \(11x+11y-x^2-xy\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(x+y\right)\left(11-x\right)\)

b) \(x^2-xy-8x+8y\)

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

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

c) \(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

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

d) \(x^2+2xy+y^2-xz-yz\)

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

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

B3) a) x(x-5)-4(x-5)=0

<=> (x-4)(x-5)=0

TH1 :x-4=0

<=.x=4

TH2 : x-5=0

<=>x=5

b) x(x-6)-7x-42=0

<=>x(x+6)-7(x+6)=0

<=>(x-7)(x+6)=0

th1;x-7=0

<=>x=7

th2; x+6=0

<=>x=-6

c)x^3-5x^2+x-5=0

<=>  x(x^2+1)-5(x^2+1)=0

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

th1:x-5=0

<=>x=5

TH2 : x^2+1=0

<=> x^2=-1 ( vo li )

=> th2 ko tồn tại 

nho thick nha  

Bài 3

a, x(x-5)-4(x-5)=0

 (x-4)(x-5)=0

=>\(\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

b,x(x+6)-7(x+6)=0

(x-7)(x+6)=0\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)

c,x^2(x-5)+(x-5)=0

(x^2+1)(x-5)=0

\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)

Lời giải:

a. $xy(x+y)-y(x+y)^2+y^2(x-y)$

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

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

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

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

$=y^2(-2y)=-2y^3$

b.

$x(x+y)^2-y(x+y)^2+xy-x^2$

$=[x(x+y)^2-y(x+y)^2]-(x^2-xy)$

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

$=(x-y)[(x+y)^2-x]=(x-y)(x^2+2xy+y^2-x)$

a: \(xy\left(x+y\right)-y\left(x+y\right)^2+y^2\left(x-y\right)\)

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

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

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

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

b: \(x\left(x+y\right)^2-y\left(x+y\right)^2+xy-x^2\)

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

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

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

\(a,x-xy+y-y^2\\=(x-xy)+(y-y^2)\\=x(1-y)+y(1-y)\\=(1-y)(x+y)\\---\\b,x^2-4x-y+4(?)\\---\\c,x^2-2x-3\\=x^2+x-3x-3\\=x(x+1)-3(x+1)\\=(x+1)(x-3)\)

Bạn xem lại đề câu b nhé!

`x-xy+y-y^2`

`=x(1-y)+y(1-y)`

`=(1-y)(x+y)`

__

`x^2-4x-y+4`

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

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

Thiếu đề?

__

`x^2-2x-3`

`=x^2+x-3x-3`

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

`=x(x+1)-3(x+1)`

`=(x+1)(x-3)`

a) 11x + 11y + x² + xy

= 11.(x+y) + x.(x+y)

= (x+y).(11+x)

b) 255 + x² - 4xy + y²

= 255 + 2xy + x² -2xy + y²

= 255 + 2xy + (x-y)²

...

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.

a) \(xy+y-2x-2\)

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

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

b) \(xy+1+x+y\)

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

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

c) \(x\left(x-1\right)+y\left(x-1\right)+z\left(x-1\right)\)

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

chắc Lan Anh mới vào nên ko bik tick đúng

\(x^2-xy-8x+8y\)

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

a: \(=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)

b: \(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)

\(=\left(x^2y^2-9\right)\left(x^2y^2-7\right)\)

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

c: \(=x^2-8x+x-8\)

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

\(=\left(x-8\right)\left(x+1\right)\)

\(a,xy+y^2-x-y\)

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

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

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

\(---\)

\(b,\left(x^2y^2-8\right)^2-1\)

\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)

\(=\left[\left(xy\right)^2-9\right]\left(x^2y^2-7\right)\)

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

\(---\)

\(c,x^2-7x-8\)

\(=x^2+x-8x-8\)

\(=\left(x^2+x\right)-\left(8x+8\right)\)

\(=x\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(x-8\right)\)

\(Toru\)