Bài 5:1/2+1/3+...+1/2008).x=2009/1+2010/2+2011/3+5016/2008-2008
(1+1/2+1/3+...+1/2008).X=2009/1+2010/2+2011/3+...+5016/2008-2008)
giúp mình vs mình đag cần gấp
\(\frac{2009}{1}+\frac{2010}{2}+...+\frac{5016}{2008-2008}\)
\(=\frac{2009}{1}+\frac{2010}{2}+...+\frac{5016}{0}\)
Sau đó QĐM(bạn tự QĐ nha)
\(=\frac{0}{0}+\frac{0}{0}+...+\frac{5016}{0}\)
\(=\frac{5016}{0}=0\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right).x=0\)
Mà \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\ne0\)
\(\Rightarrow x=0\)
a) 2010/1+2009/2+2008/3+ ... +1/2010+2010 : 1+1/2+1/3+ ... +1/2010=
b) 1/2011+1/2010+1/2009+ ... +1/3+1/2 : 2010/1+2009/2+2008/3+ ... +1/2010=
Tìm x, biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right).x=\frac{2009}{1}+\frac{2010}{2}+\frac{2011}{3}+...+\frac{4016}{2008}-2008\)
a, \(\dfrac{5}{2}-3\left(\dfrac{1}{3}-x\right)=\dfrac{1}{4}-7x\)
b, \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2008}\right).x=\dfrac{2009}{1}+\dfrac{2010}{2}+...+\dfrac{5016}{2008}-2008\)
c, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2001}{2003}\)
GIÚP VỚI , MIK CẦN GẤP
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
a, \(\dfrac{5}{2}-3\left(\dfrac{1}{3}-x\right)=\dfrac{1}{4}-7x\)
b, \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2008}\right).x=\dfrac{2009}{1}+\dfrac{2010}{2}+...+\dfrac{5016}{2008}-2008\)
c, \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{16}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2001}{2003}\)
GIÚP VỚI , MIK CẦN GẤP
(2008 x 2009 x 2010 x 2011) x (1 + 1/2 : 3/2 - 4/3)
(2008 x 2009 x 2010 x 2011) x (1 + 1/2 : 3/2 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (1 + 1/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (4/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x 0
=0
kết ban đi
đáp số: 0
dễ như ăn cháo
thảo mai ăn hủ tiếu mì nha
ai đó giúp mk với mk xin chân thành cảm ơn! a=(2010+2010/2+2009/3+2008/4+...+1/2011/ 1/2+1/3+...+1/2011) / (1/2+1/3+1/4+1/5+...+1/2009+1/2010+1/2011)
x-1 / 2013 + x-2 / 2012 + x-3 / 2011 = x-4 / 2010 + x-5 / 2009 + x-6 / 2008
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
=>x-2014=0
hay x=2014
\(B=\frac{\frac{2008}{2011}+\frac{2009}{2010}+\frac{2010}{2009}+\frac{2011}{2008}+\frac{2012}{503}}{\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}}\)