Ta có: (2x-1)²⁰¹⁸≥0 ; (y-2/5)²⁰¹⁸≥0 ; |x+y-z|≥0

=>\(\hept{\begin{cases}\left(2x-1\right)^{2018}=0\\\left(y-\frac{2}{5}\right)^{2018}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)

Chúc bạn học tốt!

Ta có : 

\(\left(2x-1\right)^{2018}\ge0\)

\(\left(y-\frac{2}{5}\right)^{2018}\ge0\)

\(\left|x+y-z\right|\ge0\)

Mà \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}+\left|x+y-z\right|=0\) ( Giả thiết ) 

\(\Rightarrow\)\(\hept{\begin{cases}\left(2x-1\right)^{2018}=0\\\left(y-\frac{2}{5}\right)^{2018}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)

Vậy \(x=\frac{1}{2}\)\(;\)\(y=\frac{2}{5}\) và \(z=\frac{9}{10}\)

Chúc bạn học tốt ~ 

 Ko biết Anh gì ơi

Ta có: \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\)

=>\(\frac{yz + zx + xy}{xyz}=0\)

=>xy+yz+xz=0

=>yz=-xy-xz; xy=-xz-yz; xz=-xy-yz

\(x^2 + 2yz = x^2 + yz + yz = x^2 - x(y + z) + yz\)

\(=x^2-xy-xz+yz\)

\(=x(x-y)-z(x-y)=(x-y)(x-z)\)

Chứng minh tương tự, ta sẽ có:

\(y^2 + 2zx = (y - z)(y - x)\)

\(z^2 + 2xy = (z - x)(z - y)\)

Đặt \(A = \frac{1}{x^2 + 2yz} + \frac{1}{y^2 + 2zx} + \frac{1}{z^2 + 2xy}\)

\(=\frac{1}{(x - y)(x - z)}+\frac{1}{(y - z)(y - x)}+\frac{1}{(z - x)(z - y)}\)

\(=\frac{-1}{(x - y)(z - x)}+\frac{-1}{(y - z)(x - y)}+\frac{-1}{(z - x)(y - z)}\)

\(=\frac{-(y - z) - (z - x) - (x - y)}{(x - y)(y - z)(z - x)}\)

=0

=>\(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}=0\)

\(\left(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}\right)\left(x^{2016}+y^{2017}+z^{2018}\right)\)

\(=0\left(x^{2016}+y^{2017}+z^{2018}\right)\)

=0

=xy+yz+xz

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