Giải
\(\frac{-7}{10^{2005}}+\frac{-15}{10^{2016}}=\frac{-7.10}{10^{2005}.10}+\frac{-15}{10^{2006}}=\frac{-70}{10^{2006}}+\frac{-15}{10^{2006}}=\frac{-85}{10^{2006}}\left(1\right).\)
\(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-15.10}{10^{2005}.10}+\frac{-7}{10^{2006}}=\frac{-150}{10^{2006}}+\frac{-7}{10^{2006}}=-\frac{157}{10^{2006}}\left(2\right).\)
Từ (1) và ( 2 ) => (1) > (2)
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