\(\Leftrightarrow\hept{\begin{cases}x+\frac{19}{5}=0\\y+\frac{1890}{1979}=0\\z-2007=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1979}\\z=2007\end{cases}}}\)

Xét đẳng thức , ta thấy :

\(\left|x+\frac{3}{4}\right|\ge0\)

\(\left|y-\frac{1}{5}\right|\ge0\)

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

=> \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|\ge0\)

Mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\) (đề bài)

=> \(\hept{\begin{cases}\left|x+\frac{3}{4}\right|=0\\\left|y-\frac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{1}{5}\\z=-\left(-\frac{3}{4}+\frac{1}{5}\right)=\frac{11}{20}\end{cases}}\)

Ta thấy một điều phê phê thế này :v  : |a| >= 0 
=> x+3/4=0 
y-1/5=0
x+y+z=0 
=> x=-3/4 =>y=1/5 => z= 3/4 - 1/5 = 11/20 
còn Trường hợp >0 Loại vì lúc ấy phương trình vô nghiệm rồi :v

\(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1979}\right|+\left|z-2007\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{19}{5}\right|=0\\\left|y+\frac{1890}{1979}\right|=0\\\left|z-2007\right|=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+\frac{19}{5}=0\\y+\frac{1890}{1979}=0\\z-2007=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-1890}{1979}\\z=2007\end{cases}}\)

\(\left(x-3\right);\left(y+x\right)\in\left\{-1;1;-7;7\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(2;-9\right);\left(4;3\right);\left(-4;3\right);\left(10;-9\right)\right\}\left(x,y\in Z\right)\)

a)\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1980}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1980}{1975}\\z=2004\end{matrix}\right.\)

b) \(\left|\dfrac{3}{4}+x\right|+\left|-\dfrac{1}{5}+y\right|+\left|x+y+z\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|\dfrac{3}{4}+x\right|=0\\\left|-\dfrac{1}{5}+y\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)

\(x=\frac{-5}{3};y=3;z=\frac{5}{3}\)

k mk nha

Vì \(\left|x+\frac{19}{5}\right|\ge0\) với \(\forall x\)

\(\left|y+\frac{1890}{1975}\right|\ge0\) với \(\forall y\)

\(\left|z-2004\right|\ge0\)với \(\forall z\)

\(\Rightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{19}{5}\right|=0\\\left|y+\frac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1975}\\z=2004\end{cases}}\)

Theo đề bài, ta có:

x(x + y + z) = -5; y(x + y + z) = 9; z(x + y + z) = 5

=> (x + y + z)(x + y + z) = -5 + 9 + 5 = 9

=> (x + y + z)² = 9

=> x + y + z \(\in\){3; -3}

Với x + y + z = 3, ta có:

   x = -5 : 3 = \(\frac{-5}{3}\)

   y = 9 : 3 = 3

   z = 5 : 3 = \(\frac{5}{3}\)

Với x + y + z = -3, ta có:

   x = -5 : (-3) = \(\frac{5}{3}\)

   y = 9 : (-3) = -3

   z = 5 : (-3) = \(\frac{-5}{3}\)

Vậy x = \(\frac{-5}{3}\); y = 3 ; z = \(\frac{5}{3}\) hoặc x = \(\frac{5}{3}\); y = -3 ; z = \(\frac{-5}{3}\).