Question

1. rút gọn

a)x^8+x^6 +x^6+x^5+x^4+x^3+x^2+x+1/x^3 -1

b)x^5+x+1/x^3+x^2+x

y^3-x^3/x^3-3x^2y+3xy^2-y^3

Answer 1

a) \(\dfrac{x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^3-1}\)

\(=\dfrac{\left(x^8+x^7+x^6\right)+\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)}{x^3-1}\)

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

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

\(=\dfrac{x^6+x^3+1}{x-1}\)

b) \(\dfrac{x^5+x+1}{x^3+x^2+x}\)

\(=\dfrac{x^5+x^4+x^3+x^2-x^4-x^3-x^2+x+1}{x^3+x^2+x}\)

\(=\dfrac{\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)}{x^3+x^2+x}\)

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

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

\(=\dfrac{x^3-x^2+1}{x}\)