K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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

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

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

11 tháng 9 2017

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x.\left(x+1\right)}=1\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x.\left(x+1\right)}=1\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=1\)

11 tháng 9 2017

MK chưa làm xong đợi mk ăn cơm xong làm nha

8 tháng 6 2016

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)

\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)

\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(-\frac{1}{x+1}=\frac{2001}{4006}-\frac{1}{2}\)

\(-\frac{1}{x+1}=-\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(\Rightarrow x=2012\)

 

 

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 \(x+2004=0\)\(\Rightarrow x=0-2004=-2004\)

12 tháng 7 2016

quy dong TS tat ca len 2 
2/6+2/12+2/20+...+2/x(x+1)
=2/2.3+2/3.4+2/4.5+...+2/x.(x+1)
=1/2-1/3+1/3-1/4+1/4-1/5+....+1/x-1/x+1
=1/2-1/x+1=1999/2001