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DB
3
VN
Võ Ngọc Phương
VIP
8 tháng 8 2023
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
E
6
N
27 tháng 8 2017
\(A=\frac{3}{1.3}+\frac{3}{3.5}+.....+\frac{3}{19.21}\)
\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{19.21}\right)\)
\(A=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{19}-\frac{1}{21}\right)\)
\(A=\frac{3}{2}.\left(1-\frac{1}{21}\right)\)
\(A=\frac{3}{2}.\frac{20}{21}\)
\(A=\frac{10}{7}\)
TP
27 tháng 8 2017
Ta có:
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{19.21}\)
\(\Rightarrow A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)
\(\Rightarrow A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{21}\right)=\frac{2}{3}.\frac{20}{21}=\frac{40}{63}\)
HT
27 tháng 10 2020
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
TN
3
22 tháng 4 2017
A= 2/1x3 + 2/3x5 + 2/5x7 +... + 2/2003x2005
A= 1 - 1/3 +1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2003 + 1/2005
A= 1 - 1/2005
A= 2004/2005
B= 2006/2005
suy ra A < B
DD
6 tháng 8 2015
\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
PN
3
2 tháng 7 2018
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{2017.2019}\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\frac{2018}{2019}\)
\(=\frac{1009}{673}\)
2 tháng 7 2018
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}.....+\frac{3}{2017.2019}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2017.2019}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\frac{2018}{2019}=\frac{1009}{673}\)
HH
0
AH
Akai Haruma
Giáo viên
31 tháng 12 2022
Lời giải:
$2\times A=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{19\times 21}$
$2\times A=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+...+\frac{21-19}{19\times 21}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{21}$
$=1-\frac{1}{21}=\frac{20}{21}$
$\Rightarrow A=\frac{20}{21}: 2= \frac{10}{21}$
VB
1
24 tháng 10 2021
\(\dfrac{8}{1\cdot3}+\dfrac{8}{3\cdot5}+...+\dfrac{8}{89\cdot91}\)
\(=4\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{89}-\dfrac{1}{91}\right)\)
\(=4\cdot\dfrac{90}{91}=\dfrac{360}{91}\)
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(\Leftrightarrow A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Leftrightarrow A=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(\Leftrightarrow A=\frac{3}{2}.\frac{100}{101}\)
\(\Leftrightarrow A=\frac{150}{101}\)
A=3/1x3+3/3x5+3/5x7+.....+3/99x101
A=3x(1/1x3+1/3x5+1/5x7+.....+1/99x101)
A=3/2x(2/1x3+2/3x5+2/5x7+.....+2/99x101)
A=3/2x(1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)
A=3/2x(1/1-1/101)
A=3/2x(101/101-1/101)
A=3/2x100/101
A=150/101.
Vậy A=150/101