Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}\)
\(=\left(\frac{3}{29}\cdot\frac{29}{3}\right)-\left(\frac{1}{5}\cdot\frac{29}{3}\right)\)
\(=1-\frac{29}{15}\)
\(=\frac{-14}{15}\)
b)\(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
=\(=\frac{16\cdot\left(-5\right)\cdot54\cdot56}{15\cdot14\cdot24\cdot21}\)
\(=\frac{2^4\cdot\left(-5\right)\cdot2\cdot3^3\cdot2^3\cdot7}{3\cdot5\cdot7\cdot2\cdot2^3\cdot3\cdot7}\)
\(=2^4\)
c)\(\frac{37}{7}\cdot\frac{8}{11}+\frac{37}{7}\cdot\frac{5}{11}-\frac{37}{7}\cdot\frac{2}{11}\)
\(=\frac{37}{7}\cdot\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)
\(=\frac{37}{7}\cdot1\)
\(=\frac{37}{7}\)
Đúng nhớ k nhen!
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
\(S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}\)
\(S>\frac{4}{10}=\frac{2}{5}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9.10}< S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{8.9}\)
=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}< S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{8}-\frac{1}{9}\)
=> \(\frac{1}{2}-\frac{1}{10}< S< 1-\frac{1}{9}\)
=> \(\frac{2}{5}< S< \frac{8}{9}\)(dpcm )
Sửa đề : Chứng minh : S > 1
Ta thấy : \(\frac{5}{20}>\frac{5}{21}>\frac{5}{22}>\frac{5}{23}>\frac{5}{24}\)
\(\Rightarrow S=\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>\frac{5}{24}\times5=\frac{25}{24}>1\)
Vậy S > 1 (ĐPCM)