\(a,A=5-3\left(2x-1\right)^2\le5\left(vì3\left(2x-1\right)^2\ge0\forall xnên-3\left(2x-1\right)^2\le0\right)\\ Dấu"="xảyrakhi:\\ 3\left(2x-1\right)^2=0\\ \Leftrightarrow x=\frac{1}{2}\\ Vậy.....\)
b,
\(B=\frac{1}{2\left(x-1\right)^2+3}\le\frac{1}{0+3}=\frac{1}{3}\left(vì2\left(x-1\right)^2\ge0\forall x\right)\\ Dấu"="xảyrakhi:\\ 2\left(x-1\right)^2=0\\ \Leftrightarrow x=1\\ Vậy...\)
c,
\(C=\frac{x^2+8}{x^2+2}=1+\frac{6}{x^2+2}\le1+\frac{6}{0+2}=4\left(vìx^2\ge0\forall x\right)\\ Dấu"="xảyrakhi:\\ x^2=0\Leftrightarrow x=0\\ Vậy......\)