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1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)-1/x=1/2010
1/x(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)-1/x=1/2010
1/x(x+1)+1/(x+1)-1/(x+3)-1/x=1/2010
-1/x+1 +(x+3)-(x+1)/(x+1)(x+3)=1/2010
-1/x+3=1/2010
x+3=-2010
x=-2013
Sx=(x+x+x+...+x)+(1+2+3+...+2010)=2010
Sx=2011x+(2010+1).2010:2=2010
Sx=2011x+2021055=2010
2011x=(-2019045)
x=(-1004.000497).
Vậy x=(-1004.000497).
1/x+x+1+x+2+x+3+...+x+2006+2007=2007
------------------------------------------=2007-2007
------------------------------------------=0
x+x+x+...+x+1+2+3+...+2006=0
2007.x+(1+2+...+2006)=0
2007.x+(2006+1).[(2006-1)+1]:2=0
2007.x+2013021=0
2007.x=0-2013021
x=-2013021:2007
x=-1003
2/x+x+1+x+2+...+x+198=401-201-200-199
199.x+(1+2+...+198)=-199
199.x+(1+198).[(198-1)+1]:2=-199
199.x+19701=-199
199.x=-199-19701
x=-19900:199
x=-100
3/x+x+1+x+2+...+x+2008=2010-2010-2009
2009.x+(2008+1).[(2008-1)+1]:2=-2009
2009.x+2017036=-2009
2009.x=-2009-2017036
x=-2019045:2009
x=-1005
\(\Leftrightarrow2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2010}{4024}=\dfrac{1005}{2012}\)
=>1/x+1=-251/1006
=>x+1=-1006/251
=>x=-1257/251
theo đề bài ta có
1+ 1/3 +1/6+...+2/x(x+1)=1+2009/2010
=>1/3+1/6+....+2/x(x+1)=2009/2010
=>1/2(2+1):2+1/3(3+1):2+.....+1/x(x+1):2=2009/2010
=>2/2(2+1)+2(3+1)+....+2/x(x+1)=2009/2010
=>2(1/2.3+1/3.4+....+1/x(x+1)=2009/2010
=>1/2-1/3+1/3-1/4+.....+1/x-1/x+1=2009/2010:2
=>1/2-1/x+1=2009/4020
=>1/x+1=1/2-2009/4020
=>1/x+1=1/4020
=>x+1=4020
=>x=4020-1
=>x=4019
A=\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right)...\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
A=\(\frac{2009}{2010}.\frac{2008}{2010}...0.\frac{-1}{2010}\)
A=0
bnaj phải viết bao nhiêu x-1 chứ(đề thiếu)
Đề bài kiểu gì đây ???