tính nhanh
A=2/1.3+2/3.5+2/5.7+...+2/49.51
B=2/5+2/7-1/6-1/8+11/35-34/48+11/12
K
Khách
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23 tháng 3 2022
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
H
15 tháng 5 2017
Có:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(1-0-0-0...-0-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{5}{11}:\dfrac{1}{2}=\dfrac{10}{11}\)
\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{10}{11}\)
\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{11}\)
\(\Rightarrow x+2=11\)
\(\Rightarrow x=11-2=9\)
Vậy x = 9.
Chúc bạn học tốt!
15 tháng 5 2017
1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2)
= 1/2.(1/1 - 1/3) + 1/2.(1/3 - 1/5) + 1/2.(1/5 - 1/7) + ... + 1/2.(1/x -1/x+2)
= 1/2.(1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 )
= 1/2.(1/1 - 0 - 1/x+2 )
= 1/2 . ( 1/1 - 1/x+2 )
= 1/2 . ( x+2/x+2 - 1/x+2 )
= 1/2 . x+1/x+2
Mà 1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2) = 5/11
=> 1/2 . x+1/x+2 = 5/11
=> x+1/x+2 = 5/11 : 1/2
=> x+1/x+2 = 10/11
=> x+1/x+2-1 = 10/11-1
=> x+1/x+1 = 10/10
=> x + 1 = 10
=> x = 10 - 1
=> x = 9
Vậy x = 9
TT
1
TT
1
HT
27 tháng 8 2015
\(\left[\frac{12}{11}-\left(\frac{1}{2}+\frac{1}{44}\right)\right].\left(x-0,2\right)=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{25}{44}.\left(x-0,2\right)=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{9.11}\right)\)
\(x-0,2=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right):\frac{25}{44}\)
\(x-\frac{1}{5}=\frac{22}{25}.\left(1-\frac{1}{11}\right)=\frac{22}{25}.\frac{10}{11}=\frac{4}{5}\)
\(x=\frac{4}{5}+\frac{1}{5}\)
\(x=1\)
A=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
=\(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{49}-\frac{2}{51}\)
= \(2.(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51})\)
=2.\((1-\frac{1}{51})\)
=\(2.\frac{50}{51}\)
=\(\frac{100}{51}\)