Có \(x^2+x-1=\left(\dfrac{-1-\sqrt{5}}{2}\right)^2+\dfrac{-1-\sqrt{5}}{2}-1\)
\(=\dfrac{6+2\sqrt{5}}{4}+\dfrac{-3-\sqrt{5}}{2}\)\(=\dfrac{3+\sqrt{5}}{2}+\dfrac{-3-\sqrt{5}}{2}\)
\(=0\)
\(\Rightarrow x^2=1-x\) và \(-x^2-x=-1\)
Có \(M=\left(4x^5+4x^4-5x^3+2x-2\right)^{2018}+2019\)
\(=\left[4x^3\left(x^2+x-1\right)-x^3-2\left(1-x\right)\right]^{2018}+2019\)
\(=\left[-x.x^2-2x^2\right]^{2018}+2019\)
\(=\left[-x\left(1-x\right)-2x^2\right]^{2018}+2019\)
\(=\left(-x^2-x\right)^{2018}+2019\)
\(=\left(-1\right)^{2018}+2019\)
\(=2020\)
Đặt \(4x^5+4x^4-5x^3+2x-2=A\)
Xét pt \(t^2+t-1=0\)
\(\Delta=b^2-4ac=1+4=5\Rightarrow\left[{}\begin{matrix}t=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{5}}{2}\\t=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)
\(\Rightarrow x\) là nghiệm của pt \(t^2+t-1=0\Rightarrow x^2+x-1=0\)
\(A=4x^5+4x^4-4x^3-x^3-x^2+x+x^2+x-1-1\)
\(=4x^3\left(x^2+x-1\right)-x\left(x^2+x-1\right)+x^2+x-1-1=-1\)
\(\Rightarrow M=\left(-1\right)^{2018}+2019=2020\)