Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)
\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)
\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)
\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)
\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)
\(A=\frac{907,425}{28561}-\frac{19}{37}\)
\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)
\(A=\frac{-509084,275}{1056757}=-0,04604282...\)
Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!
Chứng tỏ\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.........+\frac{1}{100^2}\) <\(\frac{1}{2}\)
Giờ tớ đặt cụm cần chứng minh là A
Ta có:
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
........................
\(\frac{1}{100^2}< \frac{1}{99.100}\)
=>\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+..........+\frac{1}{100^2}\) <\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.........+\frac{1}{99.100}\)
=> A < \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+.......+\frac{1}{99}-\frac{1}{100}\)
=> A <\(\frac{1}{2}-\frac{1}{100}\)
=> A<\(\frac{50}{100}-\frac{1}{100}\)
=> A<\(\frac{49}{100}\) <\(\frac{50}{100}\) =\(\frac{1}{2}\)
=> A<\(\frac{1}{2}\)
\(\frac{2010.2011-1005}{2010.2010+1005}\)
\(=\frac{2010.\left(2010+1\right)-2005}{2010.2010+1005}\)
\(=\frac{2010.2010+2010-1005}{2010.2010+1005}\)
\(=\frac{2010.2010+1005}{2010.2010+1005}\)
\(=1\)
