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HV
4
5 tháng 8 2016
Ta có: \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;...;\left|x+\frac{1}{110}\ge0\right|\)
=> \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{100}\right|\ge0\)
=> 11x \(\ge\)0 => x\(\ge\)0
=> \(x+\frac{1}{2}>0;x+\frac{1}{6}>0;...;x+\frac{1}{110}>0\)
=> \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};...;\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\)
=> \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{110}\right)=11x\)
=> 10x + \(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)
=> 10x + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)\)= 11x
=> 10x + \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)= 11x
=> 10x + \(\frac{10}{11}\)= 11x
=> x = \(\frac{10}{11}\)
Vậy x = \(\frac{10}{11}\)
SS
2
DP
14 tháng 7 2017
Sửa đề : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+....+\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{1}-\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{50}\)
\(\Leftrightarrow x=50-1=49\)
14 tháng 7 2017
Sửa đề: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
(=) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
(=)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{100}\)
(=)\(1-\frac{1}{x+1}=\frac{98}{100}\)
(=)\(\frac{1}{x+1}=1-\frac{98}{100}\)
(=)\(\frac{1}{x+1}=\frac{1}{50}\)=> \(x+1=50\)
\(x=50-1\)
\(x=49\)
T_i_c_k cho mình nha,thanks you so much!
DT
1
TV
1 tháng 4 2016
Ta có S=1/2^2+1/3^2+1/4^2+...+1/9^2
<1/2²+1/2*3+1/3*4+....+1/8*9
=1/2²+1/2-1/3+1/3-1/4+....+1/8-1/9
=1/4+1/2-1/9=23/36<32/36=8/9 (♪)
Ta lại có S=1/2^2+1/3^2+1/4^2+...+1/9^2
>1/2²+1/3*4+1/4*5+....+1/9*10
=1/2²+1/3-1/4+1/4-1/5+........+1/9-1/10
=1/2²+1/3-1/10
=19/20>8/20=2/5 ( ♫)
Từ (♪)( ♫) cho ta đpcm
NV
0
NM
5
DP
3 tháng 6 2017
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x}=\frac{99}{100}\)
Đặt \(x=n.\left(n+1\right)\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n.\left(n+1\right)}=\frac{99}{100}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{n+1-n}{n.\left(n+1\right)}=\frac{99}{100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{\left(n+1\right)}=\frac{99}{100}\)
\(=1-\frac{1}{\left(n+1\right)}=\frac{99}{100}\)
\(\frac{1}{\left(n+1\right)}=1-\frac{99}{100}=\frac{1}{100}\)
\(\Rightarrow x=\left(100-1\right).100\)
\(=9900\)
3 tháng 6 2017
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x}=\frac{99}{100}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x}=\frac{99}{100}\)
Đề kì z
LE
1
26 tháng 4 2016
=>1/1.2+1/2.3=1/3.4+........+1/x.(x+1)=2008/2009
=>1-1/2+1/2-1/3+.....+1/x-1/x+1=1-1/2009
=>1-1/x+1=1-1/2009
=>-1/x=-1/2009
=>1/x=1/2009
=>x=2009
Nhớ k cho mình nha
LT
2
29 tháng 7 2017
1/1x2+1/2x3+1/3x4+...+1/x=99?100
1/1-1/2+1/2-1/3+1/3-1/4+...+1/x=99/100
1/1-1/x=99/100
1/x=1/1-99/100
1/x=1/100
=>x=100
kbn nha