回復 1# whzzthr 的帖子
\(\begin{align}
& f\left( x \right)=\left( x-11 \right)\left( x-21 \right)\left( x-31 \right)q\left( x \right)+r\left( x \right) \\
& r\left( 11 \right)=f\left( 11 \right)=2012 \\
& r\left( 21 \right)=f\left( 21 \right)=-2013 \\
& r\left( 31 \right)=f\left( 31 \right)=2014 \\
& r\left( x \right)=g\left( x \right)=\frac{\left( x-21 \right)\left( x-31 \right)}{\left( 11-21 \right)\left( 11-31 \right)}\times 2012+\frac{\left( x-11 \right)\left( x-31 \right)}{\left( 21-11 \right)\left( 21-31 \right)}\times \left( -2013 \right)+\frac{\left( x-11 \right)\left( x-21 \right)}{\left( 31-11 \right)\left( 31-21 \right)}\times 2014 \\
& \\
& g\left( 41 \right)=2012+6039+6042=14093 \\
\end{align}\)
大略描出\(\left( 11,2012 \right),\left( 21,-2013 \right),\left( 31,2014 \right)\)這三個點，就知道\(f\left( x \right)=0\)至少有二實根，由於它是三次，故有三實根
\(f\left( x \right)-g\left( x \right)=0\)有11、21、31這三個實根
大略描出\(\left( 11,2012 \right),\left( 21,-2013 \right),\left( 31,2014 \right)\)這三個點，就知道\(r\left( x \right)=0\)有二實根
[ 本帖最後由 thepiano 於 2015-7-19 07:31 AM 編輯 ]
