\(A=0,\left(21\right)-\left|x-0,\left(4\right)\right|\)
vì \(\left|x-0,\left(4\right)\right|\ge0\) \(\Rightarrow0,\left(21\right)-\left|x-0,\left(4\right)\right|\le0,\left(21\right)\)
vậy GTLN của A là 0,(21) khi và chỉ khi x=0,(4)
\(B=1,1\left(2\right)-\left|x+1,\left(2\right)\right|\)
vì \(\left|x+1,\left(2\right)\right|\ge0\Rightarrow1,1\left(2\right)-\left|x+1,\left(2\right)\right|\le1,1\left(2\right)\)
vâyh GTLN của B là 1,1(2) khi x=-1,(2
\(A=0,\left(21\right)-\left|x-0,\left(4\right)\right|\)
\(\Rightarrow A=\frac{21}{99}-\left|x-\frac{4}{9}\right|\)
\(\Rightarrow A=\frac{7}{33}-\left|x-\frac{4}{9}\right|\)
Ta có
\(\left|x-\frac{4}{9}\right|\ge0\)
\(\frac{7}{33}-\left|x-\frac{4}{9}\right|\le\frac{7}{33}\)
Dấu " = " xảy ra khi x = 4/9
Vậy MAXA=7 / 33 khi x = 4 / 9
b)
\(B=1,1\left(2\right)-\left|x+1,\left(2\right)\right|\)
\(\Rightarrow B=0,9+0,\left(2\right)-\left|x+1+0,\left(2\right)\right|\)
\(\Rightarrow B=\frac{9}{10}+\frac{2}{9}-\left|x+1+\frac{2}{9}\right|\)
\(\Rightarrow B=\frac{101}{90}-\left|x+\frac{11}{2}\right|\)
Ta có
\(\frac{101}{90}-\left|x+\frac{11}{2}\right|\le\frac{101}{90}\) với mọi x
Dấu " = " xảy ra khi x = - 11 / 2
Vậy MAXB=101 / 90 khi x = - 11 / 2