Question

bài 2: làm tính chia

a. (x^3-3x^2+x_3):(x-3)

b. (x^2+2x+x^2-4):(x+2)

c.(2x^3-5x^2+6x-15):(2x-5)

d. (x^3+2x^2-2x-1):(x^2+3x+1)

Answer 1

a) Ta có: \(\frac{x^3-3x^2+x-3}{x-3}\)

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

b) Ta có: \(\frac{x^2+2x+x^2-4}{x+2}\)

\(=\frac{x\left(x+2\right)+\left(x+2\right)\left(x-2\right)}{x+2}=\frac{\left(x+2\right)\left(x+x-2\right)}{x+2}=2x-2\)

c) Ta có: \(\frac{2x^3-5x^2+6x-15}{2x-5}\)

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