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Đặt S là biểu thức trên
\(\Rightarrow S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{97.99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(\frac{99}{99}-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow S=\frac{49}{99}\)
Vậy biểu thức trên có giá trị là \(\frac{49}{99}\)
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{97\times99}\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+....+\frac{1}{97\times99}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\times\frac{98}{99}\)
\(=\frac{49}{99}\)
\(=\frac{6}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-............+\frac{1}{97}-\frac{1}{99}\right).\\ \)
\(=\frac{6}{2}\left(1-\frac{1}{97}\right)\)
tới đây tính máy là ra luôn
\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{2017.2019}\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\frac{2018}{2019}\)
\(=\frac{1009}{673}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}.....+\frac{3}{2017.2019}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2017.2019}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\frac{2018}{2019}=\frac{1009}{673}\)
\(=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+..+\frac{1}{9.11}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{11}\right)\)
\(=2.\left(1-\frac{1}{11}\right)\)
\(=2.\left(\frac{11}{11}-\frac{1}{11}\right)\)
\(=2.\frac{10}{11}\)
\(=\frac{20}{11}\)
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(\Leftrightarrow A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Leftrightarrow A=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(\Leftrightarrow A=\frac{3}{2}.\frac{100}{101}\)
\(\Leftrightarrow A=\frac{150}{101}\)