\(\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+...+\dfrac{4017}{2008.2009.2010}\)
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\(\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+...+\dfrac{4017}{2008.2009.2010}\)
\(\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+...+\dfrac{4017}{2008.2009.2010}\) Rút gọn rồi tính
\(S=\dfrac{4}{1.2.3}-\dfrac{1}{1.2.3}+\dfrac{6}{2.3.4}-\dfrac{1}{2.3.4}+...+\dfrac{4018}{2008.2009.2010}-\dfrac{1}{2008.2009.2010}\)
\(=\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2008.2010}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2008.2009.2010}\right)\)
\(=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)-\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2008.2009.2010}\right)\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)
\(=\left(1-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2009.2010}\right)\)
\(=1-\dfrac{1}{2009}-\dfrac{1}{2010}+\dfrac{1}{2009.2010}\)
\(=\dfrac{1}{2010}\left(\dfrac{1}{2009}-1\right)-\left(\dfrac{1}{2009}-1\right)\)
\(=\left(\dfrac{1}{2010}-1\right)\left(\dfrac{1}{2009}-1\right)=\dfrac{2009}{2010}.\dfrac{2008}{2009}=\dfrac{1004}{1005}\)
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Cho : \(S=\dfrac{5}{1.2.3}+\dfrac{8}{2.3.4}+...+\dfrac{6026}{2008.2009.2010}\). So sánh S với 2
Tìm x:\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}-3x=\left(1.2.3+2.3.4+...+98.99.100\right).\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\right)\)
Chứng tỏ rằng: \(\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{2017}{1008.1009.1010}\) < \(\dfrac{5}{4}\)
Gọi biểu thức là \(A\). Ta có :
\(A=\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{2017}{1008.1009.1010}\)
\(A=\left(\dfrac{1.2}{1.2.3}+\dfrac{2.2}{2.3.4}+\dfrac{3.2}{3.4.5}+...+\dfrac{1008.2}{1008.1009.1010}\right)+\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{1008.1009.1010}\right)\)\(A=\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{1009.1010}\right)+\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{1008.1009}-\dfrac{1}{1009.1010}\right)\)
\(A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{1009}-\dfrac{1}{1010}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{1009.1010}\right)\)
\(A< 2.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{2}=1+\dfrac{1}{4}=\dfrac{5}{4}\)
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Cho S=\(\dfrac{5}{1.2.3}+\dfrac{8}{2.3.4}+\dfrac{11}{3.4.5}+...+\dfrac{6068}{2022.2023.2024}\)
So sánh S với 2
15 tháng 4 2023 lúc 13:47
a) dfrac{2}{1.2.3}+dfrac{2}{2.3.4}+dfrac{2}{3.4.5}+...+dfrac{2}{18.19.20}b) dfrac{4}{1.3.5}+dfrac{4}{3.5.7}+dfrac{4}{5.7.9}+...+dfrac{4}{21.23.25}c) dfrac{3}{1.2}-dfrac{5}{2.3}+dfrac{7}{3.4}-dfrac{9}{4.5}+...+dfrac{39}{19.20}-dfrac{41}{20.21}d) dfrac{8}{9}cdotdfrac{15}{16}cdotdfrac{24}{25}cdot...cdotdfrac{99}{100}cdotdfrac{120}{121}e) left(1+dfrac{7}{9}right)left(1+dfrac{7}{20}right)left(1+dfrac{7}{33}right)left(1+dfrac{7}{48}right)...left(1+dfrac{7}{180}right)Các bạn không nhất thiết phải làm h...
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a) \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)
b) \(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+\dfrac{4}{5.7.9}+...+\dfrac{4}{21.23.25}\)
c) \(\dfrac{3}{1.2}-\dfrac{5}{2.3}+\dfrac{7}{3.4}-\dfrac{9}{4.5}+...+\dfrac{39}{19.20}-\dfrac{41}{20.21}\)
d) \(\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{120}{121}\)
e) \(\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right)\left(1+\dfrac{7}{48}\right)...\left(1+\dfrac{7}{180}\right)\)
Các bạn không nhất thiết phải làm hết, làm cho nó dễ hiểu được thì càng tốt để mk vận dụng
a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=179/380
b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)
\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)
c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)
=1-1/21
=20/21
d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)
\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)
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11 tháng 4 2021 lúc 20:05
Cho :
\(A=\dfrac{5}{1.2.3}+\dfrac{8}{2.3.4}+\dfrac{11}{3.4.5}+...+\dfrac{6056}{2018.2019.2020}\)
Hãy so sánh A với 2
Tìm y:-y:dfrac{1}{2}-dfrac{5}{2}4dfrac{1}{2}Tính:N dfrac{3}{4}.dfrac{8}{9}.dfrac{15}{16}....dfrac{899}{900}.dfrac{960}{961}Sdfrac{1}{1.2.3}+dfrac{1}{2.3.4}+dfrac{1}{3.4.5}+...+dfrac{1}{10.11.12}+dfrac{1}{11.12.13}
Đọc tiếp
Tìm y:
-y:\(\dfrac{1}{2}\)-\(\dfrac{5}{2}\)=4\(\dfrac{1}{2}\)
Tính:
N = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\)....\(\dfrac{899}{900}\).\(\dfrac{960}{961}\)
S=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{10.11.12}\)+\(\dfrac{1}{11.12.13}\)
Tìm y:
-y:1/2-5/2=4+1/2
-y:1/2 = 4+1/2+5/2
-y:1/2 = 7
-y = 7.2
y = -14
Vậy y = -14
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