Question

phân tích các đa thức sau thành nhân tử

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

Answer 1

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

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

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

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

Tham khảo nhé~

Answer 2

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

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

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

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

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

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

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

Answer 3

(x+2)2-y2

mình k biết viết bình phương nhé

Answer 4

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

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

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

Answer 5

Bài làm

x² + 4x - y² + 4

= ( x² + 4x + 4 ) - y²

= ( x + 2 )² - y²

= ( x + 2 + y )( x + 2 - y )

# Học tốt # 

Answer 6

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

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

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

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