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\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}=3\)
\(\Rightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)=0\)
\(\Rightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}=0\)
\(\Rightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{4}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ..
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{6}\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow3x+1=4x\)
\(\Rightarrow x=1\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}>0\\x+\dfrac{1}{3}>0\\x+\dfrac{1}{6}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\\\left|x+\dfrac{1}{3}\right|=x+\dfrac{1}{3}\\\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\end{matrix}\right.\)
Thay vào ta được:
\(x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\)
Vậy...................
Chúc bạn học tốt!!!
\(\left|x\left(x+\dfrac{1}{2}\right)\right|=x\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{1}{2}\right)=-x\\x\left(x+\dfrac{1}{2}\right)=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{1}{2}\right)+x=0\\x\left(x+\dfrac{1}{2}\right)-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{1}{2}+1\right)=0\\x\left(x+\dfrac{1}{2}-1\right)=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{3}{2}\right)=0\\x\left(x-\dfrac{1}{2}\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{3}{2};0;\dfrac{1}{2}\right\}\)
Chúc bạn học tốt!!!
\(1)\left|x\left(x+\dfrac{1}{2}\right)\right|=x\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{1}{2}\right)=x\\x\left(x+\dfrac{1}{2}\right)=x\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{1}{2}\right)+x=0\\x\left(x+\dfrac{1}{2}\right)-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{1}{2}+1\right)=0\\x\left(x+\dfrac{1}{2}-1\right)=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{3}{2}\right)=0\\x\left(x-\dfrac{1}{2}\right)=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{-3}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-3}{2};0;\dfrac{1}{2}\right\}\)
\(A=\dfrac{\left(-2\right)^0+1^{2017}+\left(\dfrac{-1}{3}\right)^8.3^8}{2^{15}}=\dfrac{3}{2^{15}}\left(1\right)\)
\(B=\dfrac{6^2}{2^{16}}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{3}{2^{15}}}{\dfrac{6^2}{2^{16}}}=\dfrac{1}{6}\)
b) \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\Rightarrow\left(x-\dfrac{2}{9}\right)^3=\left[\left(\dfrac{2}{3}\right)^2\right]^3=\left(\dfrac{4}{9}\right)^3\)
\(\Rightarrow x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{2}{3}\)
giải giùm mình nha. mới thi học kì I toán mà bài này không làm được
\(\Leftrightarrow\left[{}\begin{matrix}\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=-4\\\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=4\end{matrix}\right.\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{4}=7\\\dfrac{1}{2}x-\dfrac{1}{4}=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{29}{4}\\\dfrac{1}{2}x=-\dfrac{27}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{2}\\x=-\dfrac{27}{2}\end{matrix}\right.\)