Tim x biet \(\frac{x}{744}\)=(1-\(\frac{3030}{3131}\))+(\(\frac{616161}{626262}\)-1)+(\(\frac{9292}{9393}\)-1)
K
Khách
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12 tháng 8 2019
\(\frac{|x|}{186}=\left(1-\frac{30}{31}\right)+\left(\frac{60}{61}-1\right)\)
\(\Leftrightarrow|x|=186\left(\frac{1}{31}-\frac{1}{61}\right)\)
\(\Leftrightarrow|x|=6-\frac{186}{61}\)
\(\Leftrightarrow|x|=\frac{180}{61}\)
\(\Leftrightarrow x=\pm\frac{180}{61}\)
21 tháng 9 2015
\(1-\frac{303030}{313131}+\frac{616161}{626262}-1+\frac{929292}{939393}-1=\left(1-\frac{30}{31}\right)+\left(\frac{61}{62}-1\right)+\left(\frac{92}{93}-1\right)\)
\(=\frac{1}{31}-\frac{1}{62}-\frac{1}{93}=\frac{1}{186}\)
=> \(\frac{\left|x\right|}{186}=\frac{1}{186}\)=> |x| = 1 => x = 1 hoặc x = - 1
Vậy...............
25 tháng 5 2018
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)
\(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\)\(\ne\)0 nên x + 1 = 0 \(\Rightarrow\)x = -1
DS
26 tháng 8 2016
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\frac{\left(x+1-2\right)}{2.\left(x+1\right)}=\frac{2011}{4026}\)
F
Tim x biet
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
1
7 tháng 6 2020
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2020}\)
=> \(1-\frac{2}{x+1}=\frac{2019}{2020}\)
=> \(\frac{2}{x+1}=\frac{1}{2020}=\frac{2}{4040}\)
=> x + 1 = 4040 => x = 4039
TT
1
LM
7 tháng 9 2016
a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)
b.
\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)
\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)
\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)
\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)