Bài 1 :
a/ Ta có :
\(C=\dfrac{19}{3}-\left|x+5\right|\)
Mà \(\left|x+5\right|\ge0\)
\(\Leftrightarrow C\le\dfrac{19}{3}\)
Để \(C\) đạt GTLN thì \(\left|x+5\right|\) đạt GTNN
Dấu "=" xảy ra khi :
\(\left|x+5\right|=0\)
\(\Leftrightarrow x=-5\)
Vậy GTLN của C = 19/3 khi x = -5
b/ Ta có :
\(D=\dfrac{-21}{7}-\left|4-x\right|\)
Mà \(\left|4-x\right|\ge0\)
\(\Leftrightarrow D\le\dfrac{-21}{7}\)
Để D đạt GTLN thì \(\left|4-x\right|\) đạt GTNN
Dấu "=" xảy ra khi :
\(\left|4-x\right|=0\)
\(\Leftrightarrow x=4\)
Vậy D đạt GTLN = -21/7 khi x = 4
Bài 2 :
a/ \(\left|x-\dfrac{4}{5}\right|=\dfrac{3}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{4}{5}=\dfrac{3}{4}\\x-\dfrac{4}{5}=-\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{31}{20}\\x=\dfrac{1}{20}\end{matrix}\right.\)
Vậy ...........
b/ \(6-\left|\dfrac{1}{2}-x\right|=\dfrac{2}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{2}-x\right|=\dfrac{28}{15}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=\dfrac{28}{15}\\\dfrac{1}{2}-x=-\dfrac{28}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{41}{30}\\x=\dfrac{71}{30}\end{matrix}\right.\)
Vậy ...
