a, \(A=\dfrac{2}{3}+\left|5-x\right|\)
Vì \(\left|5-x\right|\ge0\forall x\Rightarrow\dfrac{2}{3}+\left|5-x\right|\ge\dfrac{2}{3}\forall x\)
\(\Rightarrow A_{Min}=\dfrac{2}{3}\Leftrightarrow\dfrac{2}{3}+\left|5-x\right|=\dfrac{2}{3}\)
\(\Leftrightarrow\left|5-x\right|=0\Leftrightarrow5-x=0\Leftrightarrow x=5\)
Vậy, A đạt GTNN là \(\dfrac{2}{3}\Leftrightarrow x=5\)
b, \(B=5\left(x-2\right)^2+11\)
Vì \(5\left(x-2\right)^2\ge0\forall x\Rightarrow5\left(x-2\right)^2+11\ge11\forall x\)
\(\Rightarrow B_{Min}=11\Leftrightarrow5\left(x-2\right)^2+11=11\)
\(\Leftrightarrow5\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy, B đạt GTNN là \(11\Leftrightarrow x=2\)
c, \(C=0,5-\left|x-4\right|=-\left|x-4\right|+0,5\)
Vì \(\left|x-4\right|\ge0\forall x\Rightarrow-\left|x-4\right|\le0\forall x\Rightarrow-\left|x-4\right|+5\le5\forall x\)
\(\Rightarrow C_{Max}=5\Leftrightarrow-\left|x-4\right|+5=5\)
\(\Leftrightarrow-\left|x-4\right|=0\Leftrightarrow\left|x-4\right|=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy, C đạt GTLN là \(5\Leftrightarrow x=4\)
d, \(D=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|=-\left|2x+\dfrac{2}{3}\right|+\dfrac{2}{3}\)
Vì \(\left|2x+\dfrac{2}{3}\right|\ge0\forall x\Rightarrow-\left|2x+\dfrac{2}{3}\right|\le0\forall x\Rightarrow-\left|2x+\dfrac{2}{3}\right|+\dfrac{2}{3}\le\dfrac{2}{3}\forall x\)
\(\Rightarrow D_{Max}=\dfrac{2}{3}\Leftrightarrow-\left|2x+\dfrac{2}{3}\right|+\dfrac{2}{3}=0\)
\(\Leftrightarrow-\left|2x+\dfrac{2}{3}\right|=\dfrac{-2}{3}\Leftrightarrow2x+\dfrac{2}{3}=\dfrac{2}{3}\Leftrightarrow2x=0\Leftrightarrow x=0\)
Vậy, D đạt GTLN là \(\dfrac{2}{3}\Leftrightarrow x=0\)
-- còn lại làm tương tự =))
