Question

[Nữ hoàng sến súa là ta]

Làm phép chia: 

\(a,\left(x^2+2xy+y^2\right):\left(x+y\right)\)

\(b,\left(125x^3+1\right):\left(5x+1\right)\)

\(c,\left(2x^3+5x^2-2x+3\right):\left(2x^2-x+1\right)\)

Answer 1

[kudo shinichi]

\(\frac{x^2+2xy+y^2}{x+y}=\frac{\left(x+y\right)^2}{x+y}=x+y\)

\(\frac{125x^3+1}{5x+1}=\frac{\left(5x\right)^3+1}{5x+1}=\frac{\left(5x+1\right)\left(25x^2-5x+1\right)}{5x+1}=25x^2-5x+1\)

\(\frac{2x^3+5x^2-2x+3}{2x^2-x+1}=\frac{\left(2x^3-x^2+x\right)+\left(6x^2-3x+3\right)}{2x^2-x+1}\)

\(=\frac{x\left(2x^2-x+1\right)+3.\left(2x^2-x+1\right)}{2x^2-x+1}=\frac{\left(2x^2-x+1\right)\left(x+3\right)}{2x^2-x+1}=x+3\)

Tham khảo nhé~