LM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
AL
1
12 tháng 2 2017
Lời giải đây bn nhé :
\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{89}{\left(44.45\right)^2}\)
=\(\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{89}{1936.2025}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{1936}-\frac{1}{2025}\)
=\(1-\frac{1}{2025}\)
=\(\frac{2024}{2025}\)
xong r nhé
HP
24 tháng 12 2016
1,tổng quát: (2k+1)/[k(k+1)^2]
=(2k+1)/k^2(k+1)^2=[(k+1)^^2-k^2]/k^2(k+1)^2=1/k^2-1/(k+1)^2
áp dụng vào ,kết quả =2024/2025
GH
0
VH
3
DN
19 tháng 12 2016
Ta có \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}\)
\(\Rightarrow A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow A=2\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=2\left(\frac{2016}{2017}\right)\)
\(\Rightarrow A=\frac{4032}{2017}\)
PM
19 tháng 12 2016
Ta có:\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2016\cdot2017}\)
\(=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+....+\frac{2}{2016}-\frac{2}{2017}\)
\(=\frac{2}{1}-\frac{2}{2017}=2-\frac{2}{2017}=\frac{4034}{2017}-\frac{2}{2017}=\frac{4032}{2017}\)
16 tháng 8 2016
\(\frac{2}{1.2}+\frac{2}{2.3}+..........+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(\Rightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{x\left(x+1\right)}\right)=\frac{4028}{2015}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{x}-\frac{1}{x+1}=\frac{4028}{2015}:2\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow x+1=2015\Rightarrow x=2014\)
OP
16 tháng 8 2016
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}\right)=1\frac{2013}{2015}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=1\frac{2013}{2015}\div2\)
\(1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\frac{1}{x+1}=1-\frac{2014}{2015}\)
\(\frac{1}{x+1}=\frac{1}{2015}\)
\(x+1=2015\)
\(x=2015-1\)
\(x=2014\)
KK
8 tháng 2 2017
\(C1:\)\(S\)\(=225\)\(cm^2\)\(\Leftrightarrow\)\(S=\left(4x-1\right)^2\)
\(\Rightarrow\left(4x-1\right)^2=225\)
\(\Rightarrow\left(4x-1\right)^2=15^2\Rightarrow4x-1=15\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
NV
0
TT
1
13 tháng 2 2017
\(A=\frac{3}{\left(1\cdot2\right)^2}+\frac{5}{\left(2\cdot3\right)^2}+\frac{7}{\left(3\cdot4\right)^2}+...+\frac{89}{\left(44\cdot45\right)^2}\)
\(=\frac{2^2-1^2}{1^2\cdot2^2}+\frac{3^2-2^2}{2^2\cdot3^2}+\frac{4^2-3^2}{3^2\cdot4^2}+...+\frac{45^2-44^2}{44^2\cdot45^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+....+\frac{1}{44^2}-\frac{1}{45^2}\)
\(=1-\frac{1}{45^2}=1-\frac{1}{2025}=\frac{2024}{2025}\)
28 tháng 12 2015
2/1.2+2/2.3+2/3.4+...+2/x(x+1)=4028/2015
2(1/1.2+1/2.3+1/3.4+...+1/x(x+1))=4028/2015
2(1/1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/x+1)=4028/2015
2(1-1/x+1)=4028/2015
1-1/x+1=2014/2015
(x+1-1)/x+1=2014/2015
x/x+1=2014/2015
(x+1).2014=2015x
2014x-2015x=-2014
-x=-2014
x=2014
\(\frac{1}{n^2\left(n+1\right)^2}=\frac{1}{2n+1}.\left[\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\right]\)
\(A_n=\frac{2n+1}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\\ \)
\(A=1-\frac{1}{\left(45\right)^2}\)