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HM
0
TH
30 tháng 8 2018
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}\)
\(\Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}\)
\(\Rightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}\)
\(\Rightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4020}{2011}\)
\(\Rightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{4020}{2011}:2\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2010}{2011}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2010}{2011}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2010}{2011}\)
\(\Rightarrow\dfrac{1}{x+1}=-\dfrac{2009}{4022}\)
\(\Rightarrow4022=-2009\left(x+1\right)\)
\(\Rightarrow4022=-2009x-2009\)
\(\Rightarrow2009x=-2009-4022\)
\(\Rightarrow2009x=-6031\)
\(\Rightarrow x=-\dfrac{6031}{2009}\)
NK
2
19 tháng 12 2017
(*) <=> 1\6 + 1\12 +.. + 1\x.(x+1) = 2009\(2011.2)
ma
1\2.3 =1\2-1\3
1\3.4=1\3-1\4
...............
1\x(x+1)= 1\x-1\(x+1)
cong tung ve ta dc
Vt= 1\2- 1\(x+1) =2009\(2.2011)
<=> 2011\(2.2011) -2009\(2.2011) =1\(x+1)
<=> 1\2011 =1\(x+1)
=> x=2010
19 tháng 12 2017
1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 1999/2001
nhân 1/2 vào 2 vế ta được vế trái là :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{x-1}{\left(x+1\right)}=\frac{1999}{2001}\)
suy ra : 2001x - 2001 = 1999x + 1999
2x = 1999 + 2001 = 4000
=> x = 2000
PT
0
NT
2
HT
4 tháng 9 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)
=> x + 1 = 2003
=> x = 2003 - 1
=> x = 2002
18 tháng 4 2023
13+16+110+...+1x(x+1):2=2001200313+16+110+...+1�(�+1):2=20012003
26+212+220+...+2x(x+1)=2001200326+212+220+...+2�(�+1)=20012003
2.(12.3+13.4+14.5+...+1x(x+1))=200120032.(12.3+13.4+14.5+...+1�(�+1))=20012003
12−13+13−14+14−15+...+1x−1x+1=20012003:212−13+13−14+14−15+...+1�−1�+1=20012003:2
12−1x+1=2001400612−1�+1=20014006
=> 1x+1=12−20014006=120031�+1=12−20014006=12003
=> x + 1 = 2003
=> x = 2003 - 1
=> x = 2002
8 tháng 6 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(-\frac{1}{x+1}=\frac{2001}{4006}-\frac{1}{2}\)
\(-\frac{1}{x+1}=-\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(\Rightarrow x=2012\)
NM
8 tháng 6 2016
Ta có: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}:2\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{2003}{4006}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2003}{4006}-\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4006}=\frac{1}{2003}\)
=> x + 1 = 2003
=> x = 2002
Vậy x = 2002
Duyệt nha !!!
chúc hk tốt!!!
5 tháng 6 2017
Sorry mink mới lớp 5 nên ko thể giúp bn lm bài toán này thành thật xin lỗi
5 tháng 6 2017
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Dễ thấy \(\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}\)nên biểu thức trong ngoặc thứ hai \(\ne\)0
Do đó \(x+1=0\)\(\Rightarrow x=0-1=-1\)
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+4}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}\)nên biểu thức trong ngoặc thứ hai phải \(\ne\)0
Do đó \(x+2004=0\)\(\Rightarrow x=0-2004=-2004\)
HA
1
14 tháng 7 2015
1/3+1/6+1/10+...+1/x*(2x+1)=1999/2001
2/6+2/12+...2/x(x+1)=1999/2001
2[1/2*3+1/3*4+...+1/x(x+1)]=1999/2001
1/2-1/3+1/3-1/4+...+1/x-1/x+1=1999/2001:2
(1/2-1/x+1)+(1/3-1/3)+...+(1/x-1/x)=1999/4002
1/2-1/x+1=1999/4002
1/x+1=1/2-1999/4002
1/x+1=1/2001
=>(x+1)=2001
x=2001-1
x=2000
Vậy x=2000
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x.\left(x+1\right)}=1\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x.\left(x+1\right)}=1\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=1\)
MK chưa làm xong đợi mk ăn cơm xong làm nha