a) \(A=4x-x^2+3\)
\(\Leftrightarrow A=-\left(x^2-4x+4\right)+7\)
\(\Leftrightarrow A=-\left(x-2\right)^2+7\le7,\forall x\in R\)
\(\Rightarrow GTLN\left(A\right)=7\left(tại.x=2\right)\)
b) \(B=-3x^2+5x+2\)
\(\Leftrightarrow B=-3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+\dfrac{25}{12}+2\)
\(\Leftrightarrow B=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\le\dfrac{49}{12},\forall x\in R\)
\(\Rightarrow GTLN\left(B\right)=\dfrac{49}{12}\left(tại.x=\dfrac{5}{6}\right)\)
\(a,A=4x-x^2+3\)
\(=-\left(x^2-4x+4\right)+7\)
\(=-\left(x-2\right)^2+7\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2+7\le7\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow-\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(Max_A=7\) khi \(x=2\)
\(b,B=-3x^2+5x+2\)
\(=-3x^2+5x-\dfrac{25}{12}+\dfrac{25}{12}+2\)
\(=-3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+\dfrac{49}{12}\)
\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\)
Ta có: \(\left(x-\dfrac{5}{6}\right)^2\ge0\forall x\)
\(\Rightarrow-3\left(x-\dfrac{5}{6}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\le\dfrac{49}{12}\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)
Vậy \(Max_B=\dfrac{49}{12}\) khi \(x=\dfrac{5}{6}\)
#Toru
