a) \(\left|1-x\right|+\left|y-\frac{2}{3}\right|+\left|x+z\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}1-x=0\\y-\frac{2}{3}=0\\x+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1-0=1\\y=0+\frac{2}{3}=\frac{2}{3}\\z=0-1=-1\end{cases}}}\)

Vậy \(x=1,y=\frac{2}{3},z=-1\)

b) \(\left|\frac{1}{4}-x\right|+\left|x+y+z\right|+\left|\frac{2}{3}+y\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{4}-x=0\\x+y+z=0\\\frac{2}{3}+y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}-0=\frac{1}{4}\\x+y+z=0\\y=0+\frac{2}{3}=\frac{2}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{4}\\z=0-\frac{1}{4}-\frac{2}{3}=\frac{-11}{12}\\y=\frac{2}{3}\end{cases}}}\)

Vậy \(x=\frac{1}{4},y=\frac{-11}{12},z=\frac{2}{3}\)

x(x+y+z) + y(x+y+z) + z(x+y+z) = 2 + 25 - 2 = 25 

=> ( x+ y+ z )(x+y+z) = 25 

=> x + y+ z = 5 hoặc x + y +z = -5 

(+) x + y +z = 5 => x.5 = 2 => x = 2/5 

                        => y.5=5 => y = 1 

                        => z.5 = -2 => z = -2/5 

(+) x+ y+ z = -5 => -5x = 2 => x= -2/5 (loại x > 0)

Vậy x = 2/5 ; y = 1 ; z = -2/5 

cộng 3 vế của đẳng thức , ta được :

x . ( x + y + z ) + y . ( x + y + z ) + z . ( x + y + z ) = 2 + 25 + ( -2 )

= ( x + y + z ) . ( x + y + z ) = 25

= ( x + y + z )² = 5²

\(\Rightarrow\)x + y + z = 5

\(\Rightarrow\)x = \(\frac{2}{5}\); y = 5 ; z = \(\frac{-2}{5}\)

Nếu x + y + z = -5 thì sao

a) Vì : \(\left|x+\frac{19}{5}\right|\ge0\forall x\in R\)

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

\(\left|z-2004\right|\ge0\forall z\in R\)

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

Dấu''='' xảy ra khi và chỉ khi \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-1890}{1975}\\z=2004\end{cases}}\)

b,\(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)

Ta có:\(\left|x+\frac{9}{2}\right|\ge0\forall x\)

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

          \(\left|z+\frac{7}{2}\right|\ge0\forall z\)

\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\forall x,y,z\)

Mà \(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)

\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+\frac{9}{2}=0\\y+\frac{4}{3}=0\\z+\frac{7}{2}=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{9}{2}\\y=-\frac{4}{3}\\z=-\frac{7}{2}\end{cases}}}\)

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

Ta có:\(\left|x+\frac{3}{4}\right|\ge0\forall x\)

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

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

\(\Rightarrow\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

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

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

hreury