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\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{20}\)
\(=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)\)
\(>\frac{1}{15}\cdot5+\frac{1}{20}\cdot5\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}\)
Bài làm
Ta có:
\(\frac{1}{11}>\frac{1}{20}\), \(\frac{1}{12}>\frac{1}{20}\), \(\frac{1}{13}>\frac{1}{20}\), \(\frac{1}{14}>\frac{1}{20}\), \(\frac{1}{15}>\frac{1}{20}\), \(\frac{1}{16}>\frac{1}{20}\), \(\frac{1}{17}>\frac{1}{20}\), \(\frac{1}{18}>\frac{1}{20}\),\(\frac{1}{19}>\frac{1}{20}\)
=> \(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}\)
hay \(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}\)
=> \(S=\frac{1}{20}.10=\frac{10}{20}=\frac{1}{2}\)
Do đó: \(S=\frac{1}{2}\)
# Chúc bạn học tốt #
\(\left(1+\frac{1}{4}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right).\left(1+\frac{1}{24}\right)...\left(1+\frac{1}{9999}\right)\)
\(=\frac{5}{4}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{10000}{9999}=\frac{5.9.16.25...10000}{4.8.15.24...9999}=\frac{5.3^2.4^2.5^2...100^2}{4.2.4.3.5.4.6...99.101}\)
\(=\frac{5.3.4.5...100.3.4.5...100}{4.2.3.4...99.4.5.6...101}=\frac{5.100.3}{4.2.101}=\frac{5.25.3}{2.101}=\frac{375}{202}.\)
a)\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)vaB=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
+)Ta có:\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)\)
\(\Leftrightarrow A=\frac{31}{23}-\left(\frac{7}{32}+\frac{128}{32}\right)\)
\(\Leftrightarrow A=\frac{31}{23}-\frac{135}{32}\)
\(\Leftrightarrow A=\frac{992}{736}-\frac{3105}{736}\)
\(\Leftrightarrow A=\frac{-2113}{736}\left(1\right)\)
+)Ta lại có:\(B=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(\Leftrightarrow B=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(\Leftrightarrow B=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(\Leftrightarrow B=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)
\(\Leftrightarrow B=\frac{1}{3}+\left(-1\right)+1\)
\(\Leftrightarrow B=\frac{1}{3}\left(2\right)\)
+)Từ (1) và (2)
\(\Leftrightarrow A< 0< B\Leftrightarrow A< B\)
Vậy A<B
b)\(\frac{200420042004}{200520052005}va\frac{2004}{2005}\)
+)Ta có \(\frac{200420042004}{200520052005}=\frac{2004.100010001}{2005.100010001}=\frac{2004}{2005}\)
\(\Leftrightarrow\frac{200420042004}{200520052005}=\frac{2004}{2005}\)
c)\(C=\frac{2020^{2006}+1}{2020^{2007}+1}vaD=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(C=\frac{2020^{2006}+1}{2020^{2007}+1}< 1\)
\(\Leftrightarrow C< \frac{2020^{2006}+1+2019}{2020^{2007}+1+2019}=\frac{2020^{2006}+2020}{2020^{2007}+2020}=\frac{2020.\left(2020^{2005}+1\right)}{2020.\left(2020^{2006}+1\right)}=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(\Leftrightarrow C< D\)
Chúc bạn học tốt
Ta có : \(A=\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
=> \(5A=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\)
Lấy 5A trừ A theo vế ta có :
5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\right)\)
4A = \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)
Đặt B = \(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
=> 5B = \(1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
Lấy 5B trừ B ta có :
=> 5B - B = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)\)
=> 4B =\(1-\frac{1}{5^{11}}\)
=> B = \(\frac{1}{4}-\frac{1}{5^{11}.4}\)
Khi đó 4A = \(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{1}{5^{12}}\)
=> A = \(\frac{1}{16}-\left(\frac{1}{5^{11}.16}+\frac{1}{5^{12}.4}\right)< \frac{1}{16}\left(\text{ĐPCM}\right)\)
cậu ơi , mình quên không ghi 1 dữ liệu ạ
n thuộc N
V ậy có cần phải chỉnh sửa ở trong bài làm không ạ?????
\(a,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(b,\frac{x}{y}=\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{3}=\frac{y}{5}\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\)
\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{18}{8}=\frac{9}{4}\)
\(\Rightarrow\frac{x}{3}=\frac{9}{4}\Rightarrow x=\frac{27}{4}\)
\(\frac{y}{5}=\frac{9}{4}\Rightarrow y=\frac{45}{4}\)
Ta có : \(\left(1\frac{1}{15}\right).\left(1\frac{1}{16}\right).....\left(1\frac{1}{2020}\right)=\frac{16}{15}.\frac{17}{16}....\frac{2021}{2020}=\frac{16.17....2021}{15.16....2020}=\frac{2021}{15}\)
thank cậu nhiều nhiều