Question

Phân tích thành nhân tử

\(16-x^2\)

`16-3x+1^2`

`x^4 y^4 +4x^2 y^2 +4`

`y^2 -4y+4-x^2`

Answer 1

\(16-x^2\)

\(=\left(4-x\right)\left(4+x\right)\)

\(---\)

\(16-3x+1^2\) (kt lại đề bài nhé)

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

\(=\left[\left(xy\right)^2\right]^2+2\cdot\left(xy\right)^2\cdot2+2^2\)

\(=\left[\left(xy\right)^2+2\right]^2=\left(x^2y^2+2\right)^2\)

\(---\)

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

\(=y^2-2\cdot y\cdot2+2^2-x^2\)

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

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