\(已知：m^2+m-1=0,求解：\frac{m^5+303}{m+60}\)
\[
\begin{align*}
m^2 + m – 1 =0 \\
& m^2 = 1-m \\
& m^3 = (1-m) \cdot m \\
& =m-m^2 \\
& =m-(1-m) \\
& = 2m-1 \\
\\
& m^5 = m^2 \cdot m^3 \\
& = (1-m)(2m-1) \\
& = 2m-1 -2m^2 +m \\
& = 3m-2m^2 -1 \\
& = 3m-2(1-m)-1 \\
& = 5m-3
\\
\\
& \frac{m^5+303}{m+60} = \frac{5m-3+303}{m+60} \\
& = 5
\end{align*}
\]