(x+1)+(x+3)+...+(x+99)=0

Tổng các số hạng là: (99+1):2=50 (số hạng)

=> (x+1)+(x+3)+...+(x+99)=0 <=> 50.x+(1+3+5+...+99) = 0

<=> 50.x+\frac{\left(99+1\right).50}{2}2(99+1).50​=0 <=> 50.x+2500=0 => x=-2500/50=-50

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 .......................................................!!!