\(A=5-\left|2x-1\right|\)
\(\left|2x-1\right|\ge0\Rightarrow5-\left|2x-1\right|\le5\)
\(M\text{ax}A=5\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
^^
\(B=\frac{1}{\left|x-2\right|+3}\)
\(\left|x-2\right|\ge0\Rightarrow\left|x-2\right|+3\ge3\Rightarrow\frac{1}{\left|x-2\right|+3}\le\frac{1}{3}\)
\(M\text{ax}B=\frac{1}{3}\Leftrightarrow x-2=0\Leftrightarrow x=2\)
a) Có: \(-\left|2x-1\right|\le0\)
=> \(5-\left|2x-1\right|\le5\)
Vậy GTLN của A là 5 khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\)
b) Có: \(\left|x-2\right|\ge0\)
=>\(\left|x-2\right|+3\ge3\)
=> \(\frac{1}{\left|x-2\right|+3}\le\frac{1}{3}\)
Vậy GTLN của B là \(\frac{1}{3}\) khi \(x-2=0\Leftrightarrow x=2\)
