\(\left(a\right):A=-x^2+4x+2=-\left(x^2-4x-2\right)\\ =-\left[\left(x^2-4x+4\right)-6\right]\\ =-\left(x-2\right)^2+6\le6\forall x\)
\(Dấu''=''\ xảy\ ra\ khi:x-2=0<=>x=2\)
\(Vậy\ GTLN\ của\ A\ là :6<=>x=2\)
\(\left(b\right):B=x-x^2+6=-\left(x^2-x-6\right)\\ =-\left[\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{25}{4}\right]\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
\(Dấu''=''\ xảy\ ra\ khi :x-1/2=0<=>x=1/2\)
\(Vậy\ GTLN\ của\ B\ là :25/4<=>x=1/2\)
\(\left(c\right):C=-4y^2+8y-12=-\left(4y^2-8y+12\right)\\= -\left[\left(4y^2-8y+4\right)+8\right]\\ =-\left(2y-2\right)^2-8\le-8\forall y\)
\(Dấu''=''\ xảy\ ra\ khi:2y-2=0<=>y=1\)
\(Vậy\ GTLN\ của\ C\ là:-8<=>y=1\)
