\(\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+.....+\dfrac{2}{99\times100}\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)=2.\dfrac{99}{100}=\dfrac{99}{50}\)
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\(1\times2+1\times2\times3+...+1\times2\times3\times...\times100\)
Cho A=\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\)
So sánh A với 1
Đặt A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
=> A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=> A = 1 - \(\dfrac{1}{100}\) = \(\dfrac{99}{100}\)
=> 1 = \(\dfrac{100}{100}\)
=> A < 1
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A = 11.2+12.3+13.4+...+199.10011.2+12.3+13.4+...+199.100
=> A = 1−12+12−13+13−14+...+199−11001−12+12−13+13−14+...+199−1100
=> A = 1 - 11001100 = 9910099100
=> 1 = 100100100100
=> A < 1
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23 tháng 7 2023 lúc 14:17
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{98\times99}+\dfrac{1}{99\times100}\)
\(\dfrac{2}{11\times13}+\dfrac{2}{13\times15}+...+\dfrac{2}{19\times21}+\dfrac{2}{21\times23}\)
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
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a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
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\(y=\frac{1\times100+2\times99+3\times98...+99\times2+100\times1}{1\times2+2\times3+3\times4+...+99\times100+100\times101}=?\)
Tính \(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{98\times99\times100}\)
sud kênh Mik ủng hộ với tên kênh là M.ichibi
kênh làm về MINECRAFT
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\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
tự tính
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\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{99\cdot99}-\frac{1}{99\cdot100}\)
\(2A=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}=\frac{4949}{9900}\Rightarrow A=\frac{4949}{19800}\)
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Tính giá trị của mỗi phân số sau:
\(E=\dfrac{11\times3^{29}-\left(3^2\right)^{15}}{2\times3^{14}\times2\times3^{14}}\)
\(G=\dfrac{5\times3^{11}+4\times3^{12}}{3^9\times5^2-3^0\times2^3}\)
\(H=\dfrac{\left(3\times4\times2^{16}\right)^2}{11\times2^{13}\times4^{11}-16^9}\)
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
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\(\frac{1\times2}{2\times3}+\frac{2\times3}{3\times4}+\frac{3\times4}{4\times5}+...+\frac{98\times99}{99\times100}\)
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
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Tìm số nguyên tố \(n\) lớn nhất để: \(\left(1\times2\times3\times...\times97\times98\right)+\left(1\times2\times3\times...\times98\times99\times100\right)⋮n\)
\(A=\left[1-\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+......+\frac{1}{98\times99\times100}\right)\right]\times\frac{14851}{19800}\)