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A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được:
\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)
\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)
=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)
\(4A>1=>A>\frac{1}{4}\)
\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right).\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}x\frac{32}{99}=\frac{32}{198}\)
bn tự rút gọn nha mk mới làm tắt đó
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\)\(\frac{2}{97.99}\)
\(A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.........+\frac{1}{97.99}\right)\)
\(A=2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=2\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(A=2.\frac{98}{99}\)
\(A=\frac{196}{99}\)
Ta có :
\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)-x=\frac{-100}{99}\)
\(\Leftrightarrow\)\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x=\frac{-100}{99}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{99}\right)-x=\frac{-100}{99}\)
\(\Leftrightarrow\)\(\frac{98}{99}-x=\frac{-100}{99}\)
\(\Leftrightarrow\)\(x=\frac{98}{99}+\frac{100}{99}\)
\(\Leftrightarrow\)\(x=\frac{198}{99}\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
Chúc bạn học tốt ~
a) \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{24\cdot25}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{25}\)
\(\Leftrightarrow\frac{4}{25}\)
a/ =1/5-1/6+1/6-1/7+1/7-1/8+...+1/24-1/25=1/5-1/25=4/25
Mấy câu còn lại thì từ từ!
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
Tự tính
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
= 2 . ( \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ ..... + \(\frac{1}{97}\)- \(\frac{1}{99}\)
= 2 . ( \(\frac{1}{3}\)- \(\frac{1}{99}\))
= 2 . \(\frac{2}{3}\)
= \(\frac{4}{3}\)
32% = \(\frac{32}{100}\)= \(\frac{8}{25}\)
\(\frac{4}{3}\)> \(\frac{8}{25}\)=> \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\)+ ..... + \(\frac{2}{97.99}\)> 32%
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(A=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}=\frac{800}{2475}\)
\(32\%=\frac{8}{25}=\frac{792}{2475}\)
\(\frac{800}{2475}>\frac{792}{2475}\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}>32\%\)
Đặt : \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
Do \(\frac{32}{99}>32\%\)nên \(A>32\%\left(đpcm\right)\)
de y \(\frac{1}{3}\)-\(\frac{1}{5}\)=\(\frac{2}{3.5}\)
tuong tu suy ra
M=\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{99}\)
M=\(\frac{1}{3}\)-\(\frac{1}{99}\)
M=\(\frac{32}{99}\)