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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 \(\dfrac{1}{180}< \dfrac{1}{64}\)
\(\dfrac{2}{2006}< \dfrac{5}{2006}\)
\(\Rightarrow\dfrac{1}{180}+\dfrac{2}{2006}< \dfrac{1}{64}+\dfrac{5}{2006}\)
\(\Rightarrow C< D\)
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cách giải là
\(\frac{4}{9}\)và \(\frac{13}{18}\)\(\Rightarrow\frac{4}{9}=\frac{4.2}{9.2}=\frac{8}{18}\)\(,\frac{13}{18}\)GIỮ NGUYÊN
VÌ \(\frac{8}{18}< \frac{13}{18}\)NÊN \(\frac{4}{9}< \frac{13}{18}\)
\(\frac{-15}{7}\)VÀ \(\frac{-6}{5}\)\(\Rightarrow\frac{-15}{7}=\frac{-15.5}{7.5}=\frac{-75}{35}\)
\(\frac{-6}{5}=\frac{-6.7}{5.7}=\frac{-42}{35}\)
VÌ \(\frac{-75}{35}< \frac{-42}{35}\) NÊN \(\frac{-15}{7}< \frac{-6}{5}\)
MK CHẮC CHẮN SẼ ĐÚNG
\(\frac{4}{9}< \frac{13}{18}\)
\(\frac{-15}{7}< \frac{-6}{5}\)
a) ta có :
\(\frac{2}{-7}=\frac{-2}{7}=\frac{-22}{77}\) ; \(\frac{-3}{11}=\frac{-21}{77}\)
vì \(\frac{-22}{77}