`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`

\(2ab^2-a^2b-b^3=b^2\left(2a-a^2-b\right)\)

\(2ab^2-a^2b-b^3\)

\(=-b\left(a^2-2ab+b^2\right)\)

\(=-b\left(a-b\right)^2\)

\(\left(a^2-b^2\right)+\left(a^3+b^3\right)-a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a-b\right)+\left(a+b\right)\left(a^2-ab+b^2\right)-a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a-b+a^2+b^2-ab-a^2b^2\right)\)

\(=\left(a+b\right)\left[b^2\left(1-a^2\right)+a\left(1+a\right)-b.\left(1+a\right)\right]\)

\(=\left(a+b\right)\left(a+1\right)\left(b^2+a-b\right)\)

Đa thức không phân tích được thành nhân tử bạn nhé. 

toán lớp 8 nhưg mk mới học lớp 7 thôi!

thông cảm! mik ko thể giúp bn đc

\(2ab^2-a^2b-b^3=-b\left(a^2-2ab+b^2\right)=-b\left(a-b\right)^2.\)

a) \(10x^2+10xy-x-y=10x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(10x-1\right)\)

\(\left(a^2+b^2+c^2\right)^2-a^2b^2-b^2c^2-a^2c^2\)

\(=a^4+b^4+a^2b^2+2a^2b^2+2a^3b+2ab^3-a^2b^2-b^2c^2-c^2a^2\)

\(=\left(a^4+2a^2b^2+b^4\right)+\left(2a^3b+2ab^3\right)-\left(a^2c^2+b^2c^2\right)\)

\(=\left(a^2+b^2\right)^2+2ab.\left(a^2+b^2\right)-c^2.\left(a^2+b^2\right)\)

\(=\left(a^2+b^2\right).\left(a^2+b^2+2ab-c^2\right)\)

\(=\left(a^2+b^2\right).\left[\left(a+b\right)^2-c^2\right]\)

\(=\left(a^2+b^2\right).\left(a+b-c\right).\left(a+b+c\right)\)