=>x+1/3 =a hoặc x+1/3=-a

=>x=a-1/3   hoặc x=-a-1/3

tíc mình nha

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

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|y-\frac{3}{4}\right|=0\\\left|z+1\right|=0\end{cases}}\)

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

\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{3}{4}\\z=-1\end{cases}}\)

a) Ta có: \(\left|x-1,5\right|+\left|2,5-x\right|=0\)

   mà \(\left|x-1,5\right|+\left|2,5-x\right|\ge\left|x-1,5+2,5-x\right|=1\)

nên ko tồn tại x 

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

\(\Leftrightarrow\begin{cases}\left|x-2\right|=0\\\left|y+\frac{1}{2}\right|=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-2=0\\y+\frac{1}{2}=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=2\\y=-\frac{1}{2}\end{cases}\)

a) Để A thuộc Z => \(\sqrt{x}\)- 3thuộc ước của 2 => \(\sqrt{x}\)- 3 thuộc -1; -2;1;2

=> căn x = 1 hoặc 2

câu b làm tương tự

Ta có : \(\hept{\begin{cases}\left|x-\frac{3}{4}\right|\ge0\forall x\\\left|\frac{2}{5}-y\right|\ge0\forall y\\\left|x-y+z\right|\ge0\forall x;y;z\end{cases}}\Leftrightarrow\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\z=-\frac{7}{20}\end{cases}}\)

Vậy x = 3/4 ; y = 2/5 ; z = -7/20 

\(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)

Ta có: \(\left|x-\frac{3}{4}\right|;\left|\frac{2}{5}-y\right|;\left|x-y+z\right|\ge0\Rightarrow\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\)

Mà \(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\\frac{3}{4}-\frac{2}{5}+z=0\Rightarrow z=\frac{-7}{20}\end{cases}}\)