\(a^{101}+b^{101}=a^{100}+b^{100}\Leftrightarrow a^{101}-a^{100}+b^{101}-b^{100}=0\)
\(\Leftrightarrow a^{100}\left(a-1\right)+b^{100}\left(b-1\right)=0\left(1\right)\)
\(a^{102}+b^{102}=a^{101}+b^{101}\Leftrightarrow a^{101}\left(a-1\right)+b^{101}\left(b-1\right)=0\left(2\right)\)
Trừ vế cho vế của (2) và (1):
\(\left(a-1\right)\left(a^{101}-a^{100}\right)+\left(b-1\right)\left(b^{101}-b^{100}\right)=0\)
\(\Leftrightarrow\left(a-1\right)a^{100}\left(a-1\right)+\left(b-1\right)b^{100}\left(b-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2.a^{100}+\left(b-1\right)^2b^{100}=0\)
Do \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\\a^{100}\ge0\\\left(b-1\right)^2\ge0\\b^{100}\ge0\end{matrix}\right.\) \(\Rightarrow\left(a-1\right)^2a^{100}+\left(b-1\right)^2b^{100}\ge0\)
Dấu "=" xảy ra khi và chỉ khi: \(\left(a;b\right)=\left(1;1\right);\left(1;0\right);\left(0;1\right);\left(0;0\right)\)
- Nếu \(\left(a;b\right)=\left(1;1\right)\Rightarrow S=1+1=2\)
- Nếu \(\left[{}\begin{matrix}\left(a;b\right)=\left(1;0\right)\\\left(a;b\right)=\left(0;1\right)\end{matrix}\right.\) \(\Rightarrow S+1+0=1\)
- Nếu \(\left(a;b\right)=\left(0;0\right)\) \(\Rightarrow S=0\)
