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C=1/2.(3/60.63+....+3/117.120)+1/1003
C=1/2.(1/60-1/63+....+1/117-1/120)+1/1003
....còn lại tự làm nha, bài còn lại cũng tương tự
Ta có:
\(C=\dfrac{2}{60.63}+\dfrac{2}{63.66}+...+\dfrac{2}{117.120}+\dfrac{2}{2006}\)
\(C=2\left(\dfrac{1}{60.63}+\dfrac{1}{63.66}+...+\dfrac{1}{117.120}\right)+\dfrac{2}{2006}\)
\(C=2.\dfrac{1}{3}\left(\dfrac{3}{60.63}+\dfrac{3}{63.66}+...+\dfrac{3}{117.120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}.\dfrac{1}{120}+\dfrac{2}{2006}\)
\(C=\dfrac{1}{180}+\dfrac{2}{2006}\)
Ta lại có:
\(D=\dfrac{5}{40.44}+\dfrac{5}{44.48}+...+\dfrac{5}{76.80}+\dfrac{5}{2006}\)
\(D=5\left(\dfrac{1}{40.44}+\dfrac{1}{44.48}+...+\dfrac{1}{76.80}\right)+\dfrac{5}{2006}\)
\(D=5.\dfrac{1}{4}\left(\dfrac{4}{40.44}+\dfrac{4}{44.48}+...+\dfrac{4}{76.80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}.\dfrac{1}{80}+\dfrac{5}{2006}\)
\(D=\dfrac{1}{64}+\dfrac{5}{2006}\)
Vì \(\dfrac{1}{180}< \dfrac{1}{64}\)
\(\dfrac{2}{2006}< \dfrac{5}{2006}\)
\(\Rightarrow\dfrac{1}{180}+\dfrac{2}{2006}< \dfrac{1}{64}+\dfrac{5}{2006}\)
\(\Rightarrow C< D\)
dở ẹt nhu cu net ma ko biet lamb tao hoc lop mau giao tao cung biet tra loi dung la ngu
Ta có:
A = \(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2016}\)
\(=2.\left(\frac{1}{60.63}+\frac{1}{63.66}+...+\frac{1}{117.120}\right)+\frac{2}{2016}\)
\(=2.\frac{1}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2016}\)
\(=\frac{2}{3}.\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2016}\)
\(=\frac{2}{3}.\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2016}\)
\(=\frac{2}{3}.\frac{1}{120}+\frac{2}{2016}\)
\(=\frac{1}{180}+\frac{2}{2016}\)
B = \(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2016}\)
\(=\frac{5}{4}.\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2016}\)
\(=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2016}\)
\(=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2016}\)
\(=\frac{5}{4}.\frac{1}{80}+\frac{5}{2016}\)
\(=\frac{1}{64}+\frac{5}{2016}\)
Vì \(\frac{1}{64}>\frac{1}{180}\) và \(\frac{5}{2016}>\frac{2}{2016}\) nên B > A
Vậy B > A
