a)
\(A=5x-x^2\)
\(A=-x^2+5x\)
\(A=-\left(x^2-5x\right)\)
\(A=-\left(x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right)\)
\(A=-\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\right]\)
\(A=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)
\(A=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\)
mà mũ chẵn luôn >= 0
\(\Rightarrow A\le\frac{25}{4}\)
Dấu '=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy,.........
b)
\(B=x-x^2\)
\(B=-x^2+x\)
\(B=-\left(x^2-x\right)\)
\(B=-\left(x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)\)
\(B=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(B=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
\(B=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\)
mà ( x - 1/2 )2 luôn lớn hơn hoặc bằng 0 với mọi x
\(\Rightarrow B\le\frac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Vậy,..........
\(C=4x-x^2+3=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left[\left(x-2\right)^2-7\right]\)
\(=-\left(x-2\right)^2+7\le7\)
Vậy \(C_{max}=7\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(D=-x^2+6x-11=-\left(x^2-6x+11\right)\)
\(=-\left(x^2-6x+9+2\right)\)
\(=-\left[\left(x-3\right)^2+2\right]\)
\(=-\left(x-3\right)^2-2\le-2\)
Vậy \(D_{max}=2\Leftrightarrow x-3=0\Leftrightarrow x=3\)
\(E=5-8x-x^2=-\left(x^2+8x-5\right)\)
\(=-\left(x^2+8x+16-21\right)\)
\(=-\left[\left(x+4\right)^2-21\right]=-\left(x+4\right)^2+21\le21\)
Vậy \(E_{max}=21\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
