Question

[Thơ Nụ =))]

Phân tích thành nhân tử

\(x^2+2x+1+4x+4\)

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

\(\dfrac{1}{2}x^2+\dfrac{1}{4}x+\dfrac{1}{4}x+1\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(x^2+2x+1+4x+4\)

\(=\left(x^2+2x+1\right)+\left(4x+4\right)\)

\(=\left(x+1\right)^2+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x+1+4\right)\)

\(=\left(x+1\right)\left(x+5\right)\)

b: Sửa đề: \(2x^3+6x^2+x^2+3x\)

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

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

\(=x\left(x+3\right)\left(2x+1\right)\)

c: \(\dfrac{1}{2}x^2+\dfrac{1}{4}x+\dfrac{1}{4}x+1\)

\(=\dfrac{1}{4}x\left(\dfrac{1}{4}x+1\right)+\left(\dfrac{1}{4}x+1\right)\)

\(=\left(\dfrac{1}{4}x+1\right)\left(\dfrac{1}{4}x+1\right)=\left(\dfrac{1}{4}x+1\right)^2\)