\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2004}\)
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3 tháng 6 2019
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
11 tháng 7 2017
Câu A
X + (X+1) + (X+3) +...+ (X+2003) = 2004
Số số hạng trong tổng 1 + 3 + ... + 2003 là
(2003 - 1) : 2 + 1 = 1002
Tổng dãy 1 + 3 + ... + 2003 là:
(1 + 2003) * 1002 : 2 = 1004004
=> (1003.X) + 1004004 = 2004
=> (1003.X)= 2004 - 1004004
=> 1003.X = - 1002000
X = - 1002000/1003
E chỉ giải đc đến đây thui!!!!!!!!!!!!!!! :)))
11 tháng 7 2017
x + ( x + 1) + (x + 3) ... + (x + 2003) = 2004
x + x + x + ... + x (có 1003 x) + 1 + 3 + 5 + ... + 2003 = 2004
x . 1003 + 1004004 = 2004
x . 1003 = 2004 - 1004004
x . 1003 = -1002000
x = -1002000 : 1003
x = -999,00299 = ~-999
1 tháng 3 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{x+1-2}{2\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow2003.2\left(x+1\right)=4010\left(x-1\right)\)
\(\Leftrightarrow4006x+4006=4010x-4010\)
\(\Leftrightarrow-4x=-8016\)
\(\Leftrightarrow x=2004\)
Vậy x = 2004
CH
1 tháng 3 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2003}{2005}\)
\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}\right).\frac{1}{2}=\frac{2003}{2005}.\frac{1}{2}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{2}{x.\left(x+1\right).2}=\frac{2003}{4020}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x.\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{x+1}{\left(x+1\right).x}-\frac{x}{\left(x+1\right).x}=\frac{2003}{4020}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4020}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4020}=\frac{7}{4020}\)
\(\frac{7}{\left(x+1\right).7}=\frac{7}{4020}\)
\(\left(x+1\right).7=4020\)
\(\Rightarrow x=....\)
16 tháng 2 2015
đặt a=1/3+1/6+1/10+...........+2/n(n+1)
1/2a=1/6+1/12+...........+1/n(n+1)
1/2a=1/2.3+1/3.4+........+1/n(n+1)
1/2a=1/2-1/3+1/3-1/4+.......+1/n-1/n+1
1/2a=1/2-1/n+1
a=(1/2--1/n+1):1/2=2003/2004
1/2-1/n+1=2003/2004.1/2
1/2-1/n+1=2003/4008
1/n+1=1/2-2003/4008
1/n+1=1/4008
suy ra n+1=4008
n=4007
L
31 tháng 5 2017
câu 2 :
\(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x + 1) = 4007/2004
2/2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1) = 4007/2004
2 × (1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x + 1)) = 4007/2004
1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1 = 4007/2004 : 2
1 - 1/x + 1 = 4007/2004 × 1/2
x/x + 1 = 4007/4008
=> x = 4007