tìm số tự nhiên x biết
a.\(\frac{1}{100}< \frac{x}{110}< \frac{1}{50}\)
b.\(\frac{123}{1000}< \frac{x}{2008}< \frac{124}{1000}\)
K
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TN
17 tháng 4 2016
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1000}{2002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1000}{2002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2002}\)
<=>x+1=2002
=>x=2001
AN
2
25 tháng 5 2018
\(\frac{x+1}{125}+\frac{x+2}{124}+\frac{x+3}{123}+\frac{x+4}{122}+\frac{x+146}{5}=0\)
\(\left(\frac{x+1}{125}+1\right)+\left(\frac{x+2}{124}+1\right)+\left(\frac{x+3}{123}+1\right)+\left(\frac{x+4}{122}+1\right)+\left(\frac{x+146}{5}-4\right)=0\)
\(\frac{x+126}{125}+\frac{x+126}{124}+\frac{x+126}{123}+\frac{x+126}{122}+\frac{x+126}{5}=0\)
\(\left(x+126\right).\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)=0\)
vì \(\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)\ne0\)nên x + 126 = 0 \(\Rightarrow\)x = -126
XO
2 tháng 2 2020
Ta có : \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|\ge0\forall x\)
=> 11x \(\ge\)0
=> x \(\ge\)0
Khi đó \(\orbr{\begin{cases}x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\left(10\text{ số hạng x }\right)\\x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=-11x\left(10\text{ số hạng x}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=11x\\10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=-11x\end{cases}}\)
=> \(\orbr{\begin{cases}10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\\10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=-11x\end{cases}}\)
=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=-11x\end{cases}}\)
=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{11}\right)=-11x\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{10}{11}\\21x=-\frac{10}{11}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{10}{11}\left(\text{tm}\right)\\x=-\frac{10}{231}\left(\text{loại}\right)\end{cases}}}\)
Vậy \(x=\frac{10}{11}\)
4 tháng 5 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009
}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x+1-1}{x+1}=\frac{2008}{2009}\)
\(\frac{x}{x+1}=\frac{2008}{2009}\)
\(2009x=2008\left(x+1\right)\)
\(2009x=2008x+2008\)
\(2009x-2008x=2008\)
\(x=2008\)
Vậy x=2008
\(\frac{1}{10}< \frac{x}{11}< \frac{1}{5}\Rightarrow\frac{11}{110}< \frac{x}{110}< \frac{22}{110}\Rightarrow x\in12;...;21\)