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