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\(C=...\)
\(=\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{4^2}+\frac{1}{5^2}-\frac{1}{5^2}+...-\frac{1}{14^2}+\frac{1}{14^2}-\frac{1}{15^2}\)
\(=\frac{1}{3^2}-\frac{1}{15^2}\)
\(=\frac{1}{9}-\frac{1}{225}\)
Do \(\frac{1}{9}-\frac{1}{225}\)<\(\frac{1}{5}\)
\(=>C< \frac{1}{5}\)( ĐPCM )
C = ...
=> C = \(\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{5^2}+...+\frac{1}{14^2}-\frac{1}{15^2}\)
C = \(\frac{1}{3^2}-\frac{1}{15^2}\)
Ta thấy : \(\frac{1}{3^2}< \frac{1}{5}\Leftrightarrow\frac{1}{3^2}-\frac{1}{15^2}< \frac{1}{5}\)
=> C < \(\frac{1}{5}\)
\(A=-\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
\(=\frac{-6.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)
\(=-\frac{6}{9}=-\frac{2}{3}\)
\(D=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
Làm tắt nha :
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(D=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{98}{100}\)
\(D=\frac{1}{2}\left(\frac{98}{99}-\frac{98}{100}\right)\)
Tự tính nốt nha
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
\(B=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\times\frac{\frac{1}{3}-0.25+0.2}{1\frac{1}{6}-0.875+0.7}+\frac{6}{7}\)\(\frac{6}{7}\)
\(B=\frac{1\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\times\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}+\frac{6}{7}\)
\(=\frac{1}{2}\times\frac{1\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\times\frac{6}{7}\)
\(=\frac{1}{2}\times\frac{1\left(\frac{2}{6}-\frac{2}{8}+\frac{2}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\times\frac{6}{7}\)
\(=\frac{1}{2}\times\frac{1\times1}{7\times2}\times\frac{6}{7}=\frac{6}{2\times7\times2\times7}=\frac{3}{98}\)
Đề có sai không bạn? Phạm Nguyễn Thục Anh
đề không sai đâu nhé
mà mình biết làm rùi