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Đặt \(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{60}\right)+...+\frac{1}{70}\)
Nhận xét:
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{30}{60}=\frac{1}{2}\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}+...+\frac{1}{70}>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}=\frac{4}{3}\)
\(\Rightarrow A>\frac{4}{3}\)
Vậy \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}>\frac{4}{3}\) (Đpcm)
\(A=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+....+\dfrac{1}{70}\\ =\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\right)+\left(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{30}\right)+\left(\dfrac{1}{30}+\dfrac{1}{31}+....+\dfrac{1}{60}\right)+....+\dfrac{1}{70}\\ \)
\(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}=\dfrac{1}{2}\)
\(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+....+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)
\(\dfrac{1}{30}+\dfrac{1}{31}+....+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{30}{60}=\dfrac{1}{2}\)
\(\Rightarrow A>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{61}+...+\dfrac{1}{70}>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{4}{3}\)
Chúc bạn học tốt !!!!!!
Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)
\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}\)
\(=\dfrac{2}{45}\)
-Chúc bạn học tốt-
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
11: \(=\dfrac{-5}{7}+\dfrac{5}{67}+\dfrac{13}{30}+\dfrac{1}{2}-\dfrac{11}{6}+\dfrac{17}{14}+\dfrac{2}{5}\)
\(=\left(\dfrac{-5}{7}+\dfrac{1}{2}+\dfrac{17}{14}\right)+\left(\dfrac{13}{30}-\dfrac{11}{6}+\dfrac{2}{5}\right)+\dfrac{5}{67}\)
\(=\dfrac{-10+7+17}{14}+\dfrac{13-55+12}{30}+\dfrac{5}{67}\)
\(=1-1+\dfrac{5}{67}=\dfrac{5}{67}\)
12: \(=\dfrac{-1}{4}\cdot\dfrac{152}{11}-\dfrac{1}{4}\cdot\dfrac{68}{11}\)
\(=\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)=-\dfrac{1}{4}\cdot20=-5\)