Theo đề ra, ta có:
\(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)
\(\Leftrightarrow\left(a^{100}+b^{100}\right).\left(a^{102}+b^{102}\right)=\left(a^{101}+b^{101}\right)^2\)
\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)+a^{202}+b^{202}=a^{202}+b^{202}+2a^{101}.b^{101}\)
\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)=2a^{101}.b^{101}\)
\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2-2ab\right)=0\)
\(\Leftrightarrow a=b=0\)
\(\Rightarrow a^{100}+b^{100}=a^{101}+b^{101}\)
\(\Rightarrow a^{100}=a^{101}\)
\(\Leftrightarrow a^{100}.\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)
\(\Rightarrow A=a^{2015}+b^{2015}=1+1=2\).
\(Từ:\) \(a^{100}+b^{100}=a^{101}+b^{101}\)
\(\Leftrightarrow a^{100}\left(a-1\right)+b^{100}\left(b-1\right)=0\left(1\right)\)
\(và\) \(a^{101}+b^{101}=a^{102}+b^{102}\)
\(\Leftrightarrow a^{101}\left(a-1\right)+b^{101}\left(b-1\right)=0 \left(2\right)\)
\(Từ\left(1\right)\) \(và\) \(\left(2\right)\)
\(\Rightarrow a^{101}\left(a-1\right)+b^{101}\left(b-1\right)-a^{100}\left(a-1\right)-b^{100}\left(b-1\right)=0\)
\(\Leftrightarrow a^{100}\left(a-1\right)^2+b^{100}\left(b-1\right)^2\)
\(Do\) \(a,b>0\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Rightarrow A=1+1=2\)
em không chắc cho lắm ạ
