Question

Phân tích đa thức thành nhân tử:

\(3x^2+6x+3-3y^2\)

Answer 1

\(3x^2+6x+3-2y^2=3x^2+3x+3x+3-3y^2\)

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

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

\(=3\left(x+1\right)^2-3y^2=3.\left[\left(x+1\right)^2-y^2\right]\)

Chúc bạn học tốt!!!

Answer 2

3x² + 6x + 3 - 3y²

= 3(x² + 2x + 1 - y²)

= \(3\left[\left(x+1\right)^2-y^2\right]\)

= \(3\left[\left(x+1+y\right)\left(x+1-y\right)\right]\)

Answer 3

3x² + 6x +3 - 3y² =(3x² + 6x +3) - 3y²

= 3(x² + 2x +1) -3y²

= 3(x+1)² - 3y²

=3\(\left[\left(x+1\right)^2-y^2\right]\)

= 3 (x+1+y)(x+!-y)

Answer 4

\(3x^2+6x+3-3y^2\)

\(\Leftrightarrow3\left(x^2+2x+1-y^2\right)\)

\(\Leftrightarrow3\left[\left(x+1\right)^2-y^2\right]\)

\(\Leftrightarrow3\left(x+1-y\right)\left(x+1+y\right)\)