Question

Phân tích đa thức thành nhân tử:

Answer 1

\(4abcd+\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(a^2c^2+b^2d^2+2abcd\right)+\left(a^2d^2+b^2c^2+2abcd\right)\)

\(=\left(ac+bd\right)^2+\left(ad+bc\right)^2\)

\(2cd\left(a^2+b^2\right)+2ab\left(c^2+d^2\right)=\left(2a^2cd+2abc^2\right)+\left(2b^2cd+2abd^2\right)\)

\(=2ac\left(ad+bc\right)+2bd\left(bc+ad\right)=2\left(ac+bd\right)\left(bc+ad\right)\)

Do đó:

\(M=\left[\left(ac+bd\right)^2+\left(ad+bc\right)^2-2\left(ac+bd\right)\left(ad+bc\right)\right]\left[\left(ac+bd\right)^2+\left(ad+bc\right)^2+2\left(ac+bd\right)\left(ad+bc\right)\right]\)

\(=\left(ac+bd-ad-bc\right)^2\left(ac+bd+ad+bc\right)^2\)

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