ta có \(a^{2012}+b^{2012}=a^{2013}+b^{2013}\)
\(\Rightarrow a^{2012}-a^{2013}+b^{2012}_{ }-b^{2013}=0\)
\(\Rightarrow a^{2012}\left(1-a\right)+b^{2012}\left(1-b\right)=0\)\(\left(1\right)\)
tương tự \(a^{2013}+b^{2013}=a^{2014}+b^{2014}\)
\(\Leftrightarrow a^{2013}\left(1-a\right)+b^{2013}\left(1-b\right)=0\)\(\left(2\right)\)
trừ (1) cho (2)
ta có \(\left(a^{2012}-a^{2013}\right)\left(1-a\right)\)\(+\left(b^{2012}-b^{2013}\right)\left(1-b\right)=0\)
\(\Leftrightarrow a^{2012}\left(1-a\right)^2+b^{2012}\left(1-b\right)^2=0\)
mà\(a^{2012}\left(1-a\right)^2\ge0;b^{2012}\left(1-b\right)^2\ge0\)
\(\Rightarrow a=1;b=1\)
\(\Rightarrow M=20\times1+11\times1+2013=2044\)
lay cai dau tru cai thu 2
xong lay cai thu 2 tru cai thu 3
xong lay ket qua dau tim dc tru ket qua sau la tim dc a=b=1
roi thay vao tinh M la xong
Ta có: \(a^{2012}+b^{2012}=a^{2013}+b^{2012}=a^{2014}+b^{2014}\)
\(\Rightarrow a^{2012}+b^{2012}-2\left(a^{2013}+b^{2013}\right)+a^{2014}+b^{2014}=0\)
\(\Rightarrow a^{2012}+b^{2012}-2\left(a^{2013}+b^{2013}\right)+a^{2014}+b^{2014}=0\)
\(\Leftrightarrow\left(a^{1006}-a^{1007}\right)^2+\left(b^{1006}-b^{1007}\right)=0\)
Từ đó ta có 2 TH
\(\hept{\begin{cases}a^{1006}-a^{1007}=0\\b^{1006}-b^{1007}=0\end{cases}\hept{\begin{cases}a=0;a=1\\b=0;b=1\end{cases}}}\)
Vậy P=20.0+11.0+2013=2013
P=20.1+11.0+2013=2033
P=20.0+11.1+2013=2024
