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\(x:\dfrac{1}{2}+x:\dfrac{1}{4}+x:\dfrac{1}{8}+...+x:\dfrac{1}{512}=511\\ 2x+4x+8x+..+512x=511\\ x\left(2+4+8+...+512\right)=511\\ x\left(2^1+2^2+2^3+...+2^9\right)=511\\ \)
Gọi \(S=2^1+2^2+2^3+...+2^9\)
\(2S=2^2+2^3+2^4+...+2^{10}\\ 2S-S=\left(2^2+2^3+2^4+...+2^{10}\right)-\left(2^1+2^2+2^3+...+2^9\right)\\ S=2^{10}-2\)
\(x\left(2^{10}-2\right)=511\\ 2x\left(2^9-1\right)=511\\ 2x\left(512-1\right)=511\\ 2x\cdot511=511\\ 2x=1\\ x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
\(x+\left|\dfrac{1}{2}-\dfrac{1}{3}\right|=\left|\dfrac{-2}{3}-\dfrac{1}{4}\right|\)
\(x+\left|\dfrac{1}{6}\right|=\left|\dfrac{-11}{12}\right|\)
\(x+\dfrac{1}{6}=\dfrac{11}{12}\)
\(x=\dfrac{11}{12}-\dfrac{1}{6}\)
\(x=\dfrac{3}{4}\)
Vậy ...
1) \(-3+\dfrac{1}{2}< x< \dfrac{7}{2}+\dfrac{1}{3}\)
\(\Rightarrow-\dfrac{5}{2}< x< \dfrac{23}{6}\)
\(\Rightarrow-2,5< x< 3,8\left(3\right)\)
Mà x là số nguyên nên:
\(x\in\left\{-2;-1;0;1;2;3\right\}\)
\(-\dfrac{2}{5}+\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{7}{6}\)
\(\Rightarrow\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\)
\(\Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\)
\(\Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\Rightarrow x=\dfrac{147}{20}\)
Chúc bạn học tốt!!!
\(\dfrac{x-1}{2}=\dfrac{8}{x-1}\)
\(\Rightarrow\left(x-1\right)\left(x-1\right)=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-1=4\Rightarrow x=5\\x-1=-4\Rightarrow x=-3\end{matrix}\right.\)