Ta có : \(a^2+b^2=c^2+d^2\)
\(\Leftrightarrow a^2-c^2=d^2-b^2\)
\(\Leftrightarrow\left(a-c\right)\left(a+c\right)=\left(d-b\right)\left(d+b\right)\)
Do \(a+b=c+d\Rightarrow a-c=d-b\)
\(\Rightarrow\left(a-c\right)\left(a+c\right)=\left(a-c\right)\left(d+b\right)\)
\(\Leftrightarrow\left(a-c\right)\left(a+c-b-d\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-c=0=d-b\\a+c=b+d\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=c\\d=b\end{matrix}\right.\\a+c=b+d\end{matrix}\right.\)
Với a = c ; d = b \(\Rightarrow a^{2012}+b^{2012}=c^{2012}+d^{2012}\left(đpcm\right)\)
Với \(a+c=b+d\)
Mà \(a+b=c+d\)
\(\Rightarrow a+c+a+b=b+d+c+d\)
\(\Rightarrow2a=2d\Rightarrow a=d\Rightarrow a^{2012}=d^{2012}\left(1\right)\)
Lại có : \(a+c=b+d\)
\(\Rightarrow b=c\Rightarrow b^{2012}=c^{2012}\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow a^{2012}+b^{2012}=c^{2012}+d^{2012}\left(đpcm\right)\)
