so sánh : N=\(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}vàM=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
so sánh \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}và\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
so sánh ko quy đồng\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}B=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
Xét A ta có
A=\(\frac{-7}{10^{2005}}\) + \(\frac{-15}{10^{2006}}\)
A=\(\frac{-7}{10^{2005}}\) +\(\frac{-8}{10^{2006}}\) +\(\frac{-7}{10^{2006}}\)
Xét B ta có
B=\(\frac{-15}{10^{2005}}\) +\(\frac{-7}{10^{2006}}\)
B=\(\frac{-8}{10^{2005}}\) + \(\frac{-7}{10^{2005}}\) +\(\frac{-7}{10^{2006}}\)
Vì \(\frac{-8}{10^{2006}}\) >\(\frac{-8}{10^{2005}}\) nên A>B
So sánh :
N = \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)và M =\(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
\(N=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)
\(M=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)
Ta xét M và N, ta có: \(\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}\text{ chung}\)
Mà: \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\Rightarrow M>N\)
N =\(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)
M=\(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
so sánh N và M
SO SÁNH \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)VÀ \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
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)
học tốt
So sánh : N = \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)và M = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
so sánh không quy đồng mẫu:\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}B=\frac{-15}{2005}+\frac{-7}{10^{2006}}\)
so sánh không qua quy đồng: \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}B=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
\(A=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)
\(B=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)
Vì \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\)
\(\Rightarrow A>B\)
A > B.
Tích nha bạn !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
hình như có nhầm chút thì phải?
-8/10^2006 < -8/10^2005 chứ!!!
\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}B=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)So sánh không quy đồng A và B
Ta có
\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)
\(B=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)
Vì \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\)
=>A>B