a) \(A=11-10x-x^2\)
\(A=-\left(x^2+10x-11\right)\)
\(A=-\left(x^2+10x+25-36\right)\)
\(A=-\left[\left(x+5\right)^2-36\right]\)
\(A=36-\left(x+5\right)^2\le36\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-5\)
b) \(B=24x-6x^2+13\)
\(B=-6\left(x^2-4x-\frac{13}{6}\right)\)
\(B=-6\left(x^2-4x+4-\frac{37}{6}\right)\)
\(B=-6\left[\left(x-2\right)^2-\frac{37}{6}\right]\)
\(B=37-6\left(x-2\right)^2\le37\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=2\)
c) \(C=-8x^2+2x-3\)
\(C=-8\left(x^2-\frac{1}{4}x+\frac{3}{8}\right)\)
\(C=-8\left(x^2-2\cdot x\cdot\frac{1}{8}+\frac{1}{64}+\frac{23}{64}\right)\)
\(C=-8\left[\left(x-\frac{1}{8}\right)^2+\frac{23}{64}\right]\)
\(C=\frac{-23}{8}-8\left(x-\frac{1}{8}\right)^2\le\frac{-23}{8}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{8}\)
\(A=11-10x-x^2=-25-10x-x^2+36=-\left(x+5\right)^2+36\le0+36=36\Rightarrow A_{max}=36\Leftrightarrow x+5=0\Leftrightarrow x=-5\)\(B=24x-6x^2+13=24x-6x^2-24+37=-6\left(x^2-4x+4\right)+37=-6\left(x-2\right)^2+37\le0+37=37\Leftrightarrow x-2=0\Leftrightarrow x=2\)C tương tự tách -8 ra ngòai
