Kỷ niệm:
2^x+5y=21
nếu x khác 0=> y=2n+1
<=>
\(2^{x-1}+5n=16\)
nếu x>1=> n=2k
\(2^{x-2}+5k=8\)
nếu x>2
\(2^{x-3}+5t=4\)
nếu x>3 \(\Rightarrow2^{x-4}+5z=2\)
nếu x>4\(\Rightarrow2^{x-5}+5p=1\)
\(\hept{\begin{cases}2^{x-5}\ge1\\2^{x-5}+5p=1\Rightarrow p=0\Rightarrow z=0\Rightarrow t=0\Rightarrow k=0\Rightarrow n=0\Rightarrow y=1\end{cases}}\)
tiếp k\(\ne1\) thì sao