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

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

= (x - y). (x + y) - 10 ( x - y)

= [( x + y) - 10)] . ( x - y)

  \(x^2\) - y² - 10\(x\) - 10y 

= (\(x^2\) - y²) - 10.(\(x\) + y)

= (\(x-y\))(\(x\) + y) - 10.(\(x\) + y)

= (\(x\) + y).(\(x\) - y - 10)

b) \(x^2y-x^3-10y+10x\)

\(=x^2\left(y-x\right)-10\left(y-x\right)\)

\(=\left(y-x\right)\left(x^2-10\right)\)

c) \(x^2\left(x-2\right)+49\left(2-x\right)\)

\(=\left(x-2\right)\left(x^2-49\right)\)

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

`#3107.101107`

`x^2 - y^2 + 10x - 10y`

`= (x^2 - y^2) + (10x - 10y)`

`= (x - y)(x + y) + 10(x - y)`

`= (x + y + 10)(x - y)`

_____

Sử dụng HĐT:

`A^2 - B^2 = (A - B)(A + B).`

Bạn xem lại đề

\(x^2-xy-10x+10y\)

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

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

\(=2xy+5x+4y^2+10y\)

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

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

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

\(1,=6xy\left(x^2-2xy+y^2\right)=6xy\left(x-y\right)^2\\ 2,=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\\ 3,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ 4,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ 5,=\left(x-1\right)^2-y^2=\left(x+y-1\right)\left(x-y-1\right)\\ 6,Sửa:x^2-x-2=x^2+x-2x-2=\left(x+1\right)\left(x-2\right)\\ 7,=x^4-4x^2-x^2+4=\left(x^2-4\right)\left(x^2-1\right)\\ =\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\\ 8,=-x^3-x^2-x=-x\left(x^2+x+1\right)\\ 9,=\left(a-3\right)\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\\ =\left(a-3\right)\left(a^2+9a+18\right)\\ =\left(a-3\right)\left(a^2+3a+6a+18\right)\\ =\left(a-3\right)\left(a+3\right)\left(a+6\right)\)

\(10,=x^2y-x^2z+y^2z-xy^2+z^2\left(x-y\right)\\ =xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\\ =\left(x-y\right)\left(xy-xz-yz+z^2\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)

x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung

=(x^5-5x^2)-(2xy+10y)

=(x^5-5x^2)-(2xy-10y)

=x^2.(x-5)-2y.(x-5)

=(x^2-2y).(x-5)

k nha

\(x^3-5x^2-2xy+10y=x^2\left(x-5\right)-2y\left(x-5\right)=\left(x-5\right)\left(x^2-2y\right)=\left(x-5\right)\left(x-2y\right)\left(x+2y\right)\)

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

\(=\left(x-5\right)\left(x^2y-2y\right)\)

xong phân tích nốt cái bậc 2 kia để được max điểm :)))