\(A=5-3.\left(2x-1\right)^2\)
Ta có: \(\left(2x-1\right)^2\ge0\) \(\forall x.\)
\(\Rightarrow-3.\left(2x-1\right)^2\le0\)\(\forall x.\)
\(\Rightarrow5-3.\left(2x-1\right)^2\le5\) \(\forall x.\)
\(\Rightarrow A\le5.\)
Dấu '' = '' xảy ra khi:
\(\left(2x-1\right)^2=0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow2x=0+1\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}.\)
Vậy \(MAX_A=5\) khi \(x=\frac{1}{2}.\)
\(D=5-\left|2x-1\right|\)
Ta có: \(\left|2x-1\right|\ge0\) \(\forall x.\)
\(\Rightarrow-\left|2x-1\right|\le0\) \(\forall x.\)
\(\Rightarrow5-\left|2x-1\right|\le5\) \(\forall x.\)
\(\Rightarrow D\le5.\)
Dấu '' = '' xảy ra khi:
\(2x-1=0\)
\(\Rightarrow2x=0+1\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}.\)
Vậy \(MAX_D=5\) khi \(x=\frac{1}{2}.\)
Chúc bạn học tốt!