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

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

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

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

toán lớp 8 nhưg mk mới học lớp 7 thôi!

thông cảm! mik ko thể giúp bn đc

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

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

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

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

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

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

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

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

e) Ta có: \(a^3-a^2-a+1\)

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

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

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

f) Ta có: \(x^3-2xy-x^2y+2y^2\)

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

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

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

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

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

d) Đề sai ko ???

e) \(a^3-a^2-a+1=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)=\left(a-1\right)^2\left(a+1\right)\)

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

Ai kb vs tui ko 

buồn quá hời!!

OLM đừng trừ điểm en nha 

đừng đưa nội quy mk bt hết oi

kiếm bn để chat thui!! ^.^

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

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

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

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

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

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

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