Question

Cho x = 1 + \(\sqrt[3]{2}+\sqrt[3]{4}\)

Tính M = \(x^5\)\(-4x^4+x^3-x^2-2x+2015\)

Answer 1

\(x=1+\sqrt[3]{2}+\sqrt[3]{4}\Rightarrow x-1=\sqrt[3]{2}+\sqrt[3]{4}\)

\(\Rightarrow\left(x-1\right)^3=2+4+3\sqrt[3]{2.4}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)=6+6\left(x-1\right)=6x\)

\(\Rightarrow x^3-3x^2+3x-1=6x\Rightarrow x^3-3x^2-3x-1=0\)

Ta có:

\(M=\left(x^5-3x^4-3x^3-x^2\right)-x^4+4x^3-2x+2015\)

\(\Rightarrow M=x^2\left(x^3-3x^2-3x-1\right)-x^4+3x^3+3x^2+x+x^3-3x^2-3x-1+2016\)

\(\Rightarrow M=-x\left(x^3-3x^2-3x-1\right)+\left(x^3-3x^2-3x-1\right)+2016\)

\(\Rightarrow M=2016\)