1) \(A=\left(2x^2+1\right)^4-3\ge0-3=-3\) (do \(\left(2x^2+1\right)^4\ge0\forall x\))
Dấu "=" xảy ra \(\Leftrightarrow\left(2x^2+1\right)=0\Leftrightarrow2x^2=-1\Leftrightarrow x^2=-\frac{1}{2}\) (vô lí)
Vậy đề sai ~v (hay là tui làm sai ta)
1b) \(B=3\left|1-2x\right|-5\ge0-5=-5\) (do \(\left|1-2x\right|\ge0\forall x\))
Dấu "=" xảy ra khi \(\left|1-2x\right|=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
Vậy \(B_{min}=-5\Leftrightarrow x=\frac{1}{2}\)
2) a) \(M=-\left|2-3x\right|-\frac{1}{2}\)
M lớn nhất khi \(-\left|2-3x\right|\) lớn nhất.
Mà \(-\left|2-3x\right|\le0\forall x\)
Nên \(M\le0-\frac{1}{2}=-\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow-\left|2-3x\right|=0\Leftrightarrow2-3x=0\Leftrightarrow x=\frac{2}{3}\)
Vậy \(M _{max}=-\frac{1}{2}\Leftrightarrow x=\frac{2}{3}\)
b) \(N=-3-\left|2x+4\right|\)
N lớn nhất khi \(\left|2x+4\right|\) bé nhất
Mà \(\left|2x+4\right|\ge0\forall x\)
Suy ra \(N\le-3-0=-3\)
Dấu "=" xảy ra \(\Leftrightarrow\left|2x+4\right|=0\Leftrightarrow x=-2\)
Vậy \(N_{max}=-3\Leftrightarrow x=-2\)
\(2,M=-\left|2-3x\right|-\frac{-1}{2}\)
Vì \(\left|2-3x\right|\ge0\Rightarrow-\left|2-3x\right|\le0\)
\(\Rightarrow A\le\frac{-1}{2}\)
Dấu \("="\) xảy ra khi \(-\left|2-3x\right|=0\)
\(\Leftrightarrow\left|2-3x\right|=0\)
\(\Leftrightarrow2-3x=0\)
\(\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
Vậy \(M_{max}=-\frac{1}{2}\Leftrightarrow x=\frac{2}{3}\)
\(2,N=-3-\left|2x+4\right|\)
Vì \(\left|2x+4\right|\ge0\Rightarrow-\left|2x+4\right|\le0\)
\(\Rightarrow N\le-3\)
Dấu \("="\) xảy ra khi \(-\left|2x+4\right|=0\)
\(\Leftrightarrow\left|2x+4\right|=0\)
\(\Leftrightarrow2x+4=0\Leftrightarrow2x=-4\Leftrightarrow x=-2\)
Vậy \(N_{max}=-3\Leftrightarrow x=-2\)