\(A=-x^2-2x+3\\ A=-x^2-2x-1+4\\ A=-\left(x^2+2x+1\right)+4\\ A=-\left(x+1\right)^2+4\\ Do\left(x+1\right)^2\ge0\forall x\\ \Rightarrow-\left(x+1\right)^2\le0\forall x\\ \Rightarrow A=-\left(x+1\right)^2+4\le4\forall x\\ \text{Dấu “=” xảy ra khi: }\\ \left(x+1\right)^2=0\\ x+1=0\\ \Leftrightarrow x=-1\\ \text{Vậy}A_{\left(Max\right)}=4\text{ khi }x=-1\)
\(B=-x^2+4x-7\\ B=-x^2+4x-4-3\\ B=-\left(x^2-4x+4\right)-3\\ B=-\left(x-2\right)^2-3\\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \Rightarrow-\left(x-2\right)^2\le0\forall x\\ \Rightarrow B=-\left(x-2\right)^2-3\le-3\forall x\\ \text{Dấu “=” xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }B_{\left(Max\right)}=-3\text{ }khi\text{ }x=2\)
\(A=-x^2-2x+3\)
\(A=-x^2-2x-1+4\)
\(A=-\left(x^2+2x+1\right)+3\)
\(A=-\left(x+1\right)^2+3\)
ta có \(-\left(x+1\right)^2\le0\)
=> \(-\left(x+1\right)^2+3\le3\)
dấu " = " xảy ra
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
vạy GTLN của A là 3 khi x = -1
