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HV
4
7 tháng 3 2018
Ta có;\(\frac{4}{1\times3}+\frac{4}{3\times5}+\frac{4}{5\times7}+....+\frac{4}{19\times21}\)
\(=2\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+....+\frac{2}{19\times21}\right)\)
\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=2\times\left(1-\frac{1}{21}\right)=2\times\frac{20}{21}=\frac{40}{21}\)
AK
7 tháng 3 2018
4/1 x 3 + 4/ 3 x 5 + 4/ 5 x 7 + ....+ 4/ 17 x 19 + 4/ 19 x 21
= 2 x ( 2/ 1 x 3 + 2/ 3 x 5 + 2/ 5 x 7 + ...+ 2/ 17 x 19 + 2/ 19 x 21 )
= 2 x ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ...+ 1/17 - 1/19 + 1/19 - 1/21 )
= 2 x ( 1 - 1/21 )
= 2 x 20/21
= 40/21
Chúc bạn học giỏi !!!
7 tháng 10 2021
\(K=\dfrac{4}{1\times3}+\dfrac{4}{3\times5}+...+\dfrac{4}{299\times301}\)
\(=2\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{299\times301}\right)\)
\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)
\(=2\times\left(1-\dfrac{1}{301}\right)=2\times\dfrac{300}{301}=\dfrac{600}{301}\)
7 tháng 10 2021
\(K=\dfrac{4}{1\cdot3}+\dfrac{4}{3\cdot5}+...+\dfrac{4}{299\cdot301}\)
\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)
\(=2\cdot\dfrac{300}{301}=\dfrac{600}{301}\)
PT
2
20 tháng 12 2015
a=1/1x2+1/2x3+....+1/99x100
a=1-1/2+1/2-1/3+....+1/99-1/100
a=1-1/100
a=99/100
b=4/1x3+4/3x5+.....+4/51x53
b=2x(2/1x3+2/3x5+....+2/51x53)
b=2x(1-1/3+1/3-1/5+...+1/51-1/53)
b=2x(1-1/53)
b=2x52/53
b=104/53
đúng tick cho mình nha
VH
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 8 2023
Bài 1:
A = \(\dfrac{1}{1\times3}\) + \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) +...+ \(\dfrac{1}{2019\times2021}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{1\times3}\) + \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+...+ \(\dfrac{2}{2019\times2021}\))
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+...+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1010}{2021}\)
CD
6
PL
2 tháng 4 2019
Dễ thôi bạn à
\(A=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{2500}{49.51}\)
\(A=\frac{1.3+1}{1.3}+\frac{3.5+1}{3.5}+\frac{5.7+1}{5.7}+...+\frac{49.50+1}{49.51}\)
\(A=\frac{1.3}{1.3}+\frac{1}{1.3}+\frac{3.5}{3.5}+\frac{1}{3.5}+\frac{5.7}{5.7}+\frac{1}{5.7}+...+\frac{49.51}{49.51}+\frac{1}{49.51}\)
\(A=1+\frac{1}{1.3}+1+\frac{1}{3.5}+1+\frac{1}{5.7}+...+1+\frac{1}{49.51}\) (có: (51 - 3) : 2 + 1 = 25 chữ số 1)
\(A=25+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{49}-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\frac{50}{51}\)
\(A=25+\frac{25}{51}\)
\(A=\frac{1300}{51}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 3 2023
A = \(\dfrac{4}{1\times3}\) - \(\dfrac{8}{3\times5}\) + \(\dfrac{12}{5\times7}\) - \(\dfrac{16}{7\times9}\) + \(\dfrac{20}{9\times11}\) - \(\dfrac{24}{11\times13}\)
A = ( \(\dfrac{1}{1}+\dfrac{1}{3}\)) - ( \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\)+ \(\dfrac{1}{7}\)) - ( \(\dfrac{1}{7}\) + \(\dfrac{1}{9}\)) +( \(\dfrac{1}{9}\)+ \(\dfrac{1}{11}\)) - (\(\dfrac{1}{11}\)+\(\dfrac{1}{13}\))
A = \(\dfrac{1}{1}+\dfrac{1}{3}\) - \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}\) - \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{11}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{12}{13}\)
TN
1
26 tháng 5 2018
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\frac{8}{9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{4}{9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{x}=1-\frac{4}{9}\)
\(\Rightarrow\frac{1}{x}=\frac{5}{9}\)
\(\Rightarrow x=\frac{1.9}{5}\)
\(\Rightarrow x=\frac{9}{5}\)
Vậy x = \(\frac{9}{5}\)
b) \(\frac{2}{3}-\frac{1}{3}.\left(x-2\right)=\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}.\left(x-2\right)=\frac{2}{3}-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}.\left(x-2\right)=\frac{5}{12}\)
\(\Rightarrow x-2=\frac{5}{12}:\frac{1}{3}\)
\(\Rightarrow x-2=\frac{5}{4}\)
\(\Rightarrow x=\frac{5}{4}+2\)
\(\Rightarrow x=\frac{13}{4}\)
Vậy x = \(\frac{13}{4}\)
_Chúc bạn học tốt_
