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\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)
\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)
\(=\dfrac{75}{101}\)
\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Tính nhanh:
a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\times\dfrac{100}{101}\)
= \(\dfrac{75}{101}\)
b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)
\(=\dfrac{100}{2}\)
\(=50\)
Chứng minh \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)
\(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)
\(A=\frac{1}{1}-\frac{1}{21}\)
\(A=\frac{20}{21}\)
\(\frac{20}{21}< 1\)
=> \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)( đpcm )
* Mình sợ sai xD *
\(A=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+...+\frac{2}{55\cdot59}\)
\(A=2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{55}-\frac{1}{59}\right)\)
\(A=2\left(\frac{1}{5}-\frac{1}{59}\right)\)
\(A=2\left(\frac{59}{295}-\frac{5}{295}\right)\)
\(A=2\cdot\frac{54}{295}\)
\(A=\frac{536}{295}\)
\(4S=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\right)\)
=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}_{ }\)
=\(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{21}-\frac{1}{23}\)
=\(\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
=> \(S=\frac{4}{25}:4=\frac{4}{25}.\frac{1}{4}=\frac{1}{25}\)
\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)
\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)
\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)
\(S\times4=\frac{1}{5}-\frac{1}{25}\)
\(S\times4=\frac{4}{25}\)
\(S=\frac{1}{25}\)
\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)
`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`
`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`
`= 1/5 - 1/25`
`= 4/25`
Đặt \(B=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+\frac{2}{13\cdot17}+....+\frac{2}{97\cdot101}\)
\(\Rightarrow2B=\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+....+\frac{4}{97\cdot101}\)
\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{97}-\frac{1}{101}\)
\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}\)
\(\Leftrightarrow B=\frac{96}{505}:2\)