Đặt \(A=5-\left|3x-4\right|\)
Ta có \(\left|3x-4\right|\ge0\) \(\forall x\)
\(\Rightarrow-\left|3x-4\right|\le0\) \(\forall x\)
\(\Rightarrow5-\left|3x-4\right|\le0+5\) \(\forall x\)
(Nếu bn ko hiểu dòng 4 thì mình giải thích ntn:
\(-\left|3x-4\right|+5\le0+5\)
hay \(5-\left|3x-4\right|\le0+5\))
Tiếp nè
\(\Rightarrow A\le5\)
\(\Rightarrow A_{max}=5\) khi \(\left|3x-4\right|=0\)
\(\Rightarrow3x-4=0\)
\(3x=4\)
\(x=\dfrac{4}{3}\)
Vậy \(A_{max}=5\) khi \(x=\dfrac{4}{3}\)
Đặt \(B=\left(4x-6\right)^{2008}+8\)
Ta có: \(\left(4x-6\right)^{2008}\ge0\) \(\forall x\)
\(\Rightarrow\left(4x-6\right)^{2008}+8\ge0+8\)
\(\Rightarrow B\ge8\)
\(\Rightarrow B_{min}=8\) khi \(\left(4x-6\right)^{2008}=0\)
\(\Rightarrow4x-6=0\)
\(4x=6\)
\(x=1,5\)
Vậy \(B_{min}=8\) khi \(x=1,5\)
Chúc bn học tốt 