Question

Phân tích đa thức thành phân tử:

\(9x^2y^2-6xy^2+y^2\)

\(x^3+27-3x\left(x+3\right)\)

\(x^2-2x-4y^2-4y\)

\(a^4-2a^2b^2+b^4\)

Answer 1

\(9x^2y^2-6xy^2+y^2=\left(3xy-y\right)^2=y^2\left(3x-1\right)^2\)

\(x^3+27-3x\left(x+3\right)=\left(x+3\right)\left(x^2-3x+9\right)-3x\left(x+3\right)=\left(x+3\right)\left(x^2-6x+9\right)\)

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

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

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