Question

Cho biểu thức \(x^2-x-1=0\)

Tính giá trị biểu thức \(Q=\frac{x^6-3x^5+3x^4-x^3+2020}{x^6-x^3-3x^2-3x+2020}\)

Answer 1

Ta có : \(Q=\frac{x^6-3x^5+3x^4-x^3+2020}{x^6-x^3-3x^2-3x+2020}\)

=> \(Q=\frac{\left(x^6-x^5-x^4\right)+\left(-2x^5+2x^4+2x^3\right)+\left(2x^4-2x^3-2x^2\right)+\left(-x^3+x^2+x\right)+\left(x^2-x-1\right)+2021}{\left(x^6-x^5-x^4\right)+\left(x^5-x^4-x^3\right)+\left(2x^4-2x^3-2x^2\right)+\left(2x^3-2x^2-2x\right)+\left(x^2-x-1\right)+2021}\)

=> \(Q=\frac{x^4\left(x^2-x-1\right)-2x^3\left(x^2-x-1\right)+2x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)+\left(x^2-x-1\right)+2021}{x^4\left(x^2-x-1\right)+x^3\left(x^2-x-1\right)+2x^2\left(x^2-x-1\right)+\left(x^2-x-1\right)+2021}\)

=> \(Q=\frac{x^4.0-2x^3.0+2x^2.0-x.0+0+2021}{x^4.0+x^3.0+2x^2.0+0+2021}\)

=> \(Q=\frac{2021}{2021}=1\)