Question

phân tích đa thức thành nhân tử:

2a2b2+2a2c2+2b2c2-a4-b4-c4

Answer 1

Đặt \(A=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4\)

\(A=-\left(a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2-2\left(ca\right)^2\right)\)

\(A=-\left(a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2+2\left(ca\right)^2-4\left(ca\right)^2\right)\)

Áp dụng hàng đẳng thức \(\left(a^2-b^2+c^2\right)=a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2+2\left(ca\right)^2\):

\(A=-\left[\left(a^2-b^2+c^2\right)^2-4\left(ca\right)^2\right]\)

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

Answer 2

2222222222222a+257222222222222222222222222222222222222222222222222222222222222222222222222222222222222222a=?