a) Ta có \(\left|x-\frac{3}{5}\right|\ge0\forall x\)
=> \(\frac{1}{2}-\left|x-\frac{3}{5}\right|\le\frac{1}{2}\forall x\)
Dâu "=" xảy ra <=> \(x-\frac{3}{5}=0\Rightarrow x=\frac{3}{5}\)
Vậy Max A = 1/2 <=> x = 3/5
b) Ta có \(\left|1,4-x\right|\ge0\forall x\)
=> B = -|1,4 - x| - 2 \(\le-2\forall x\)
Dấu "=" xảy ra <=> 1,4 - x = 0
<=> x = 1,4
Vậy Max B = -2 <=> x = -1,4
\(A=\frac{1}{2}-|x-3,5|\)
\(Ta\)\(có:|x-3,5|\ge0\forall x\)
\(\Rightarrow A=\frac{1}{2}-|x-3,5|\le\frac{1}{2}-0=\frac{1}{2}\)
\(\Leftrightarrow|x-3,5|=0\Leftrightarrow x-3,5=0\Leftrightarrow x=3,5\)
Vậy GTLN của A là \(\frac{1}{2}\)khi x=3,5
B=\(-|1,4-x|-2\)
Ta có : \(|1,4-x|\ge0\forall x\Rightarrow-|1,4-x|\le0\forall x\)
\(\Rightarrow B=-|1,4-x|-2\le0-2=-2\)
\(\Leftrightarrow|1,4-x|=0\Leftrightarrow1,4-x=0\Leftrightarrow x=1,4\)
Vậy GTLN của B là -2 khi x =1,4
a) \(A=\frac{1}{2}-\left|x-3,5\right|\)
Vì \(\left|x-3,5\right|\ge0\)\(\Rightarrow-\left|x-3,5\right|\le0\)\(\Rightarrow\frac{1}{2}-\left|x-3,5\right|\le\frac{1}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow x-3,5=0\)\(\Leftrightarrow x=3,5\)
Vậy \(maxA=\frac{1}{2}\Leftrightarrow x=3,5\)
b) \(B=-\left|1,4-x\right|-2\)
Vì \(\left|1,4-x\right|\ge0\)\(\Rightarrow-\left|1,4-x\right|\le0\)\(\Rightarrow-\left|1,4-x\right|-2\le-2\)
Dấu " = " xảy ra \(\Leftrightarrow1,4-x=0\)\(\Leftrightarrow x=1,4\)
Vậy \(maxB=-2\)\(\Leftrightarrow x=1,4\)