So sánh :
N = \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)và M =\(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
Những câu hỏi liên quan
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 : N=\(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}vàM=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
Xét N ta có :
N = \(\frac{-7}{10^{2005}}\)+ \(\frac{-15}{10^{2006}}\)
N = \(\frac{-7}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)+\(\frac{-8}{10^{2006}}\)
Xét M ta có :
M = \(\frac{-15}{10^{2005}}\)+\(\frac{-7}{10^{2006}}\)
M = \(\frac{-8}{10^{2005}}\)+\(\frac{-7}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)
Vì \(\frac{-8}{10^{2006}}\)< \(\frac{-8}{10^{2005}}\) => N < M
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ta có:M-N=\(\dfrac{-15}{10^{2005}}-\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}-\dfrac{-15}{10^{2006}}\)
=\(\dfrac{-8}{10^{2015}}\)+\(\dfrac{8}{10^{2016}}\)=8(\(\dfrac{1}{10^{2006}}-\dfrac{1}{10^{2005}}\))
Ta có:102006>102005<=>\(\dfrac{1}{10^{2006}}< \dfrac{1}{10^{2005}}\)
<=>\(\dfrac{1}{10^{2006}}-\dfrac{1}{10^{2005}}< 0\)
=>M-N<0 hay N>M
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so sánh \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}và\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
So sánh : N = \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)và M = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
SO SÁNH \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)VÀ \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
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 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
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\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}B=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)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
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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 \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\)
\(\Rightarrow A>B\)
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A > B.
Tích nha bạn !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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hình như có nhầm chút thì phải?
-8/10^2006 < -8/10^2005 chứ!!!
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