BT1: Tính
2) B=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{29.30.31}\)
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7 tháng 4 2022 lúc 14:42
Tính nhanh:a) A 2^{2010} - 2^{2009} - 2^{2008} - 2^{2007} - ... - 2 - 1b) B 20 . 8 - 33 . 64 + 17 . 8 + 9 . 16 . 8 - 11 . 128c) C ( dfrac{1}{1.2.3} + dfrac{1}{2.3.4} + ... + dfrac{1}{2009.2010.2011} ) . dfrac{2010.2011}{1010.527}
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Tính nhanh:
a) A = 2\(^{2010}\) - 2\(^{2009}\) - 2\(^{2008}\) - 2\(^{2007}\) - ... - 2 - 1
b) B = 20 . 8 - 33 . 64 + 17 . 8 + 9 . 16 . 8 - 11 . 128
c) C = ( \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + ... + \(\dfrac{1}{2009.2010.2011}\) ) . \(\dfrac{2010.2011}{1010.527}\)
a) \(A=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(A=2^{2010}\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(\text{A = 1 + 2 + . . . + 2^{2008} + 2^{2009}}\)
\(\text{⇒ 2 A = 2 + 2 2 + . . + 2^{2010}}\)
⇒ \(A=2^{2010}-1\)
⇒ \(A=2^{2010}-\left(2^{2010}-1\right)\)
⇒ \(A=1\)
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b) \(B=2072\)
c) \(\dfrac{4949}{19800}\)
Xin lỗi mình không có nhiều thời gian để giải thích trên đây á nên tạm gửi ảnh mình tạo nhé . Học tốt !
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1) Tìm x biết:
\(\left(1-\dfrac{3}{10}-x\right):\left(\dfrac{19}{10}-1-\dfrac{2}{5}\right)+\dfrac{4}{5}=1\)
2) Tính nhanh
a)\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
b)\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
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Tìm x:\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}-3x=\left(1.2.3+2.3.4+...+98.99.100\right).\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\right)\)
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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Cho A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
Chứng minh A<2
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
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}
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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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So sánh S và P:
\(S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{2009.2010.2011}\)
\(P=\dfrac{1}{2}\)
\(S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{2009.2010.2011}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{2009.2010}-\dfrac{1}{2010.2011}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2010.2011}\)
\(=\dfrac{1}{2}-\dfrac{1}{4042110}< \dfrac{1}{2}\)
\(\Rightarrow\) \(S< P\)
Vậy \(S< P\)
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