Bài 1:
a: Ta có: \(\left|x+3\right|\ge0\forall x\)
\(\left|y-2\right|\ge0\forall y\)
Do đó: \(\left|x+3\right|+\left|y-2\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi x=-3 và y=2
b: Ta có: \(\left|x+3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
c: Ta có: \(\left|-x+5\right|=\left|1-5\right|\)
\(\Leftrightarrow\left|x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Bài 16:
a: Ta có: \(\left|x\right|-x=0\)
\(\Leftrightarrow\left|x\right|=x\)
\(\Leftrightarrow x\ge0\)
b: Ta có: \(\left|x\right|+x=0\)
\(\Leftrightarrow\left|x\right|=-x\)
\(\Leftrightarrow x\le0\)
c: Ta có: \(\left|x\right|-5=-12+30\)
\(\Leftrightarrow\left|x\right|=18+5=23\)
hay \(x\in\left\{-23;-23\right\}\)
d: Ta có: \(-11-\left|x\right|=-17\)
\(\Leftrightarrow\left|x\right|=6\)
hay \(x\in\left\{6;-6\right\}\)
