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EC
0
11 tháng 7 2018
\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)
\(\frac{1}{2013}x+1+(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013})=2\)
\(\frac{1}{2013}x+1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\)
\(\frac{1}{2013}x+1+\left(1-\frac{1}{2013}\right)=2\)
\(\frac{1}{2013}x+1+1-\frac{1}{2013}=2\)
\(\frac{1}{2013}x-\frac{1}{2013}+2=2\)
\(\frac{1}{2013}.\left(x-1\right)=2-2\)
\(\frac{1}{2013}.\left(x-1\right)=0\)
=> x - 1 = 0
x = 1
N
11 tháng 7 2018
\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)
\(\frac{1}{2013}x+\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\)
\(\frac{1}{2013}x+\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\)
\(\frac{1}{2013}x+\left(1-\frac{1}{2013}\right)=2\)
\(\frac{1}{2013}x+\frac{2012}{2013}=2\)
\(\frac{1}{2013}x=2-\frac{2012}{2013}\)
\(\frac{1}{2013}x=\frac{2014}{2013}\)
\(x=\frac{2014}{2013}:\frac{1}{2013}\)
=> x=2014
LT
9 tháng 3 2020
\(\frac{1}{2013}.x+1+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{2012.2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+2-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x=\frac{1}{2013}\Rightarrow x=1\)
Vậy x=1
CHÚC CÁC EM HỌC TỐT
16 tháng 11 2018
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
16 tháng 11 2018
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
SS
6
20 tháng 4 2017
Ta goi day tren la A.Ta co:A=1/2+1/6+1/12+1/20+...+1/x(x+1)=2013/2014
A=1/(1*2)+1/(2*3)+1/(3*4)+...+1/x(x+1)=2013/2014
A=1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1.=2013/2014
A=1-1/x+1=2013/2014
Suy ra:1/x+1=1-2013/2014
1/1+1=1/2014
Suy ra x+1=2014
Suy ra x=2013
BP
1
PL
7 tháng 9 2018
Ta có : x + (x + 1) + (x + 2) + .... + (x + 2012) = 2012.2013
<=> (x + x + x + ..... + x) + (1 + 2 + .... + 2012) = 2012.2013
<=> 2013x + \(\frac{2012.2013}{2}\) = 2012.2013
<=> 2013x = 2012.2013 - \(\frac{2012.2013}{2}\)
<=> 2013x = 2025078
1/2013.x+1+1/2+1/6+1/12+...+1/2012.2013=2
1/2013.x+1+1/1.2+1/2.3+1/3.4+...+1/2012.2013=2
1/2013.x+1+1-1/2+1/2-1/3+1/3-1/4+...+1/2012-1/2013=2
1/2013.x+2-1/2013=2
1/2013.x =2-2+1/2013
1/2013.x =1/2013
=>2013.x=2013
=> x=1
\(\Rightarrow\frac{1}{2013.x}+1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2012}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013.x}+2-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013.x}=2-2+\frac{1}{2013}\)
\(\Rightarrow\frac{1}{2013.x}=\frac{1}{2013}\)
\(\Rightarrow2013.x=2013\)
\(\Rightarrow x=1\)