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

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

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

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

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

= x( x - 8 ) - y( x - 8)

= (x - y)(x - 8)

câu b hình như thiếu đề bạn ạ

a, \(x^8+x^7+1\)

= \(x^7\left(x+1\right)+1\)

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

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

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

a) \(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\) \(=x^6\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

b) \(x^4+64\)

\(=\left(x^2+8\right)^2-16x^2\)

\(=\left(x^2+8+4x\right)\left(x^2+8-4x\right)\)

2x² + 10x + 8 = 2x² + 2x + 8x + 8 = (2x² + 2x) + (8x + 8)

= 2x(x + 1) + 8(x + 1) = (x + 1)(2x + 8)

\(x^2+3y^2-4xy+10x-12y+9\)

\(=\left(x^2-xy+x\right)+9x-3xy+3y^2-12y+9\)

\(=\left(x^2-xy+x\right)+\left(9x-9y+9\right)-3xy+3y^2-3y\)

\(=\left(x^2-xy+x\right)+\left(9x-9y+9\right)-\left(3xy-3y^2+3y\right)\)

\(=x\left(x-y+1\right)+9\left(x-y+1\right)-3y\left(x-y+1\right)\\ =\left(x-y+1\right)\left(x+9-3y\right)\)

đề đúng ko

Đa thức này không phân tích được nhé bạn

bài này 1h rùi,chắc chờ tui ngủ dậy làm;

= (x+y)³ - (x+y) + xy(x+y) =

= (x+y)((x+y)² -1 +xy)) = (x+y)(x² +3xy +y² -1)

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8\right)^2+2x\left(x^2+4x+8\right)+x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)\left(x^2+4x+8+2x\right)+x\left(x^2+4x+8+2x\right)\)

\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)

\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)