a: \(\left|x+\dfrac{3}{4}\right|>=0\forall x;\left|y-1\right|>=0\forall y\)
Do đó: \(\left|x+\dfrac{3}{4}\right|+\left|y-1\right|>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=1\end{matrix}\right.\)
b: \(\left|x-\dfrac{1}{2}\right|>=0\forall x\)
\(\left|y-\dfrac{1}{3}\right|>=0\forall y\)
\(\left|x+y+z\right|>=0\forall x,y,z\)
Do đó: \(\left|x-\dfrac{1}{2}\right|+\left|y-\dfrac{1}{3}\right|+\left|x+y+z\right|>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y-\dfrac{1}{3}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\\z=-x-y=-\dfrac{5}{6}\end{matrix}\right.\)
