\(1)A=-4x^2-4x-1.\)
\(A=-\left(4x^2+4x+1\right)=-\left(2x+1\right)^2.\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\in R.\)
\(\Rightarrow-\left(2x+1\right)^2\le0\forall x\in R.\\ \Rightarrow A\le0.\)
Dấu \("="\) xảy ra \(\Leftrightarrow A=0.\Leftrightarrow-\left(2x+1\right)^2=0.\)
\(\Leftrightarrow2x+1=0.\\ \Leftrightarrow x=-\dfrac{1}{2}.\)
Vậy \(GTLN\) của \(A=0\Leftrightarrow x=-\dfrac{1}{2}.\)
\(2)A=-x^2+10x-30.\\ A=-\left(x^2-10x+25+5\right).\\ A=-\left[\left(x-5\right)^2+5\right].\\ A=-\left(x-5\right)^2-5.\)
Ta có:
\(\left(x-5\right)^2\ge0\forall x\in R.\\ \Leftrightarrow-\left(x-5\right)^2\le0.\\ \Leftrightarrow-\left(x-5\right)^2-5\le-5.\)
\(\Rightarrow A\le-5.\)
Dấu \("="\) xảy ra \(\Leftrightarrow-\left(x-5\right)^2-5=-5.\Leftrightarrow x=5.\)
Vậy \(GTLN\) của \(A=-5\Leftrightarrow x=5.\)
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