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DA
1
NN
20 tháng 4 2017
Bài này dễ ý mà, vô cùng đơn giản..........
Ta có:
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{2015}{2016}.\)
\(\dfrac{2}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)
\(1\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{2015}{2016}.\)
\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{x}-\dfrac{1}{x}\right)+\left(1-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)
\(0+0+...+0+\left(1-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)
\(1-\dfrac{1}{x+2}=\dfrac{2015}{2016}.\)
\(\dfrac{1}{x+2}=1-\dfrac{2015}{2016}.\)
\(\dfrac{1}{x+2}=\dfrac{1}{2016}.\)
\(\Rightarrow x+2=2016.\)
\(\Rightarrow x=2016-2=2014.\)
Vậy \(x=2014.\)
~ Học tốt nha bn!!! ~
Bài mik đúng thì nhớ tick mik nha!!!
21 tháng 3 2016
<=>2-2/3+2/3-2/5........+2n-2n+2<2015/2016
<=>2-2n+2<2015/2016
=>n+2=1/2016
=>n=2014
21 tháng 3 2016
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{n\left(n+2\right)}\)<\(\frac{2015}{2016}\)
VT=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{5}-\frac{1}{n+2}\)=\(1-\frac{1}{n+2}\)
Ta có:\(1-\frac{1}{n+2}=\frac{2015}{2016}\Rightarrow\)\(\frac{1}{n+2}=1-\frac{2015}{2016}\)
\(\Rightarrow\)\(\frac{1}{n+2}=\frac{1}{2016}=n+2=2016\)
\(\Rightarrow\)\(n=2014\)
Vậy\(n=2014\)
NN
0
M
1
18 tháng 3 2016
Gọi \(A=\frac{1005}{2011}\)
A=1/3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)
A=1/1.3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)
A . 2=2/1.3 + 2/3.5 + 2/5.7 +......................+2/x.(x+2)
A . 2=1/1-1/3+1/3-1/5+1/5-1/7+..............+1/x-1/x+2
A . 2=1/1+(1/3-1/3)+(1/5-1/5)+..............+(1/x-1/x)-1/x+2
A . 2=1/1-1/x+2
Suy gia:1005/2011 . 2=1/1-1/x+2
2010/2011 =1/1-1/x+2
1/x+2 =1/1-2010/2011
1/x+2 =1/2011
Suy gia:x+2=2011
x =2011-2
x =2009
9 tháng 4 2018
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
9 tháng 4 2018
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
DT
Đỗ Thanh Hải
CTVVIP
26 tháng 6 2017
Bài 2
a) Ta có
S = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
S = \(\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Vì \(\dfrac{1}{13}< \dfrac{1}{12}\)
\(\dfrac{1}{14}< \dfrac{1}{12}\)
\(\dfrac{1}{15}< \dfrac{1}{12}\)
=> \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}.3\)
Lại có
\(\dfrac{1}{61}< \dfrac{1}{60}\)
\(\dfrac{1}{62}< \dfrac{1}{60}\)
\(\dfrac{1}{63}< \dfrac{1}{60}\)
=> \(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}.3\)
=> S = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) < \(\dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)
= \(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\) = \(\dfrac{1}{2}\)
=> đpcm
DT
Đỗ Thanh Hải
CTVVIP
26 tháng 6 2017
Ta có
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{2015}{2016}\)
\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{2015}{2016}\)
\(\dfrac{1}{1}-\dfrac{1}{x+2}=\dfrac{2015}{2016}\)
\(\dfrac{1}{x+2}=\dfrac{1}{1}-\dfrac{2015}{2016}\)
\(\dfrac{1}{x+2}=\dfrac{1}{2016}\)
2016 = x + 2
x = 2016 - 2
x = 2014
Vậy x = 2014 là giá trị cần tìm
TT
3
TN
6 tháng 6 2016
\(=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)
\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)
\(2x+\frac{8}{33}=-\frac{2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
\(2x=\frac{-2024}{2017}\)
\(x=-\frac{1012}{2017}\)
6 tháng 6 2016
\(2x+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{8}{33}=\frac{-2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
Số dư dài quá. Đến đây bạn tự làm tiếp nhé
21 tháng 8 2021
Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)
\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)
\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)
Suy ra: x+2=82
hay x=80
PS
20 tháng 3 2017
a) pt => 2x-x=-25+5(chuyển vế đổi dấu) =>x=-20
b)pt=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\)=\(\frac{2016}{2017}\)
=>\(1-\frac{1}{2x+1}=\frac{2016}{2017}\)=>\(\frac{2x}{2x+1}=\frac{2016}{2017}\). Nhân chéo => x=1008
