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Ta có:
\(VT=\left|x-2017\right|+\left|2019-x\right|+\left|2018-x\right|\)
\(\Rightarrow VT\ge\left|x-2017+2019-x\right|+\left|2018-x\right|\)
\(\Rightarrow VT\ge2+\left|2018-x\right|\ge2\)
Dấu "=" xảy ra khi và chỉ khi \(x=2018\Rightarrow\) pt có nghiệm duy nhất \(x=2018\)
Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
a)\(2019-\left|x-2019\right|=x\)
\(\Rightarrow2019-x=\left|x-2019\right|\)
=>\(\left|x-2019\right|=-\left(x-2019\right)\)
=>\(x-2019\le0\)
=>\(x\le2019\)
b) Vì \(\left(2x-1\right)^{2018}\ge0\forall x\)
\(\left(y-\frac{2}{5}\right)^{2018}\ge0\forall y\)
\(\left|x+y-z\right|\ge0\forall x,y,z\)
=> \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|\ge0\forall x,y,z\)
mà \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}}\)=>\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}\)
a, Ta có:
\(\left|x-2019\right|=\orbr{\begin{cases}x-2019\ge0\Rightarrow x\ge2019\\-x+2019< 0\Rightarrow x< 2019\end{cases}}\)
Xét x<2019 thì |x-2019|=-x+2019
Khi đó: 2019-(-x+2019)=x
\(\Leftrightarrow\)-x+2019=2019-x
\(\Leftrightarrow\)-x+2019+x=2019
\(\Leftrightarrow\)0x+2019=2019
\(\Leftrightarrow\)0x=0 (thỏa mãn)
Xét 2019\(\le\)x thì |x-2019|=x-2019
Khi đó 2019-(x-2019)=x
\(\Leftrightarrow\)2019-x+2019=x
\(\Leftrightarrow\)4038-x=x
\(\Leftrightarrow\)4038=2x
\(\Leftrightarrow\)x=2019(thỏa mãn)
Vậy .......................................................!!!