tính bằng cách hợp lí
A = \(\dfrac{2018\times2017-1}{2016\times2018+2017}\)
B = \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)
K
Khách
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C
2
9 tháng 6 2023
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
\(3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
\(3A-A=\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
\(2A=3-\dfrac{1}{729}=\dfrac{2186}{729}\)
\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)
LT
Lương Thị Vân Anh
CTVHS
9 tháng 6 2023
A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
3A = \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
3A - A = ( \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )
2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)
A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)
LT
Lương Thị Vân Anh
CTVHS
26 tháng 3 2023
Đặt A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
3A = 3 + 1 + \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
3A - A = ( 3 + 1 + \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )
2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)
A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)
28 tháng 5 2022
a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)
\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)
\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)
b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)
=1/3-1/3
=0
c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=2016/2017
18 tháng 3 2018
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
28 tháng 2 2022
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\\ \Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ \Rightarrow3A-A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}-\dfrac{1}{81}-\dfrac{1}{243}-\dfrac{1}{729}\\ \Rightarrow2A=1-\dfrac{1}{729}\\ \Rightarrow2A=\dfrac{728}{729}\\ \Rightarrow A=\dfrac{364}{729}\)
NK
1
17 tháng 7 2017
\(\dfrac{1}{3}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)+\(\dfrac{1}{729}\)
=\(\dfrac{243}{729}\)+\(\dfrac{81}{729}\)+\(\dfrac{27}{729}\)+\(\dfrac{3}{729}\)+\(\dfrac{1}{729}\)
=\(\dfrac{355}{729}\)
chúc bạn học tốt ạ
23 tháng 4 2017
Dạng bài tương tự như bài này, bạn áp dụng cách làm vào làm bài của bạn nhé: Câu hỏi của Dao Dao - Toán lớp 7 | Học trực tuyến
MV
23 tháng 4 2017
\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+\dfrac{2017}{4}+...+\dfrac{2017}{2018}}{\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}}\)
Đặt \(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\) là B
\(B=\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\\ =\dfrac{2017}{1}+1+\dfrac{2016}{2}+1+...+\dfrac{1}{2017}+1-2017\\ =\dfrac{2018}{1}+\dfrac{2018}{2}+...+\dfrac{2018}{2017}-2017\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\left(2018-2017\right)\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+1\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\\ =2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2018}}{2018\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}{2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017}{2018}\)
31 tháng 5 2022
Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}=\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\)
\(=\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{a+b}\)
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(=\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{4}+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)
=>A là số hữu tỉ (ĐPCM)
$A=\dfrac{2018.2017-1}{2016.2018+2017}$
$=>A={2018.2016+2018-1}{2016.2018+2017}$
$=>A={2018.2016+2017}{2016.2018+2017}$
$=>A=1$
\(A=\dfrac{2018.2017-1}{2018.2016+2017}\)
\(A=\dfrac{2018.\left(2016+1\right)-1}{2018.2016+2017}\)
\(A=\dfrac{2018.2016+2018-1}{2018.2016+2017}\)
\(A=\dfrac{2018.2016+2017}{2018.2016+2017}=1\)
\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^7}\Rightarrow B=\dfrac{1-\dfrac{1}{2187}}{2}=\dfrac{1093}{2187}\)
Chúc bạn học tốt!!!