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Ta có:
\(\left\{{}\begin{matrix}\left(x-15\right)^2=\left|x-15\right|^2\\\left|15-x\right|=\left|x-15\right|\end{matrix}\right.\)
\(\Leftrightarrow2015\left|x-15\right|+\left|x-15\right|^2-2014\left|x-15\right|=0\)
\(\Leftrightarrow\left|x-15\right|+\left|x-15\right|^2=0\Leftrightarrow\left|x-15\right|\left(\left|x-15\right|-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=16\\x=14\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\frac{8}{15}< \frac{x}{15}< -\frac{2}{15}\)
=> \(-8< x< -2\)
=> \(x\in\left\{-7;-6;-5;-4;-3\right\}\)
Ta có : \(\frac{-8}{15}< \frac{x}{15}< \frac{-2}{15}\)
\(\Rightarrow-8< x< -2\)
\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3\right\}\)
Vậy \(x\in\left\{-7;-6;-5;-4;-3\right\}\)
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\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1990}{5}+\dfrac{x-1980}{15}\\ =>\dfrac{x-5}{1990}-1+\dfrac{x-15}{1980}-1=\dfrac{x-1990}{5}-1+\dfrac{x-1980}{15}-1\\ =>\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}-\dfrac{x-1995}{5}-\dfrac{x-1995}{15}=0\\ =>\left(x-1995\right).\left(\dfrac{1}{1990}+\dfrac{1}{1980}-\dfrac{1}{5}-\dfrac{1}{15}\right)=0\\ =>x-1995=0\\ =>x=1995\)
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15+x+x=-15
2x=-15-15
2x=-15+(-15)
2x=-30
x=-30÷2
x=-15