\(A=-x^2+6x-15\)
\(A=-x^2+2.3x-9-6\)
\(\Rightarrow-A=x^2-2.3x+9+6\)
\(-A=\left(x^2-2.3.x+3^2\right)+6\)
\(-A=\left(x-3\right)^2+6\)
\(\Rightarrow A=-\left(x-3\right)^2-6\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2-6\le-6\forall x\)
\(A=-6\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy Amax =-6\(\Leftrightarrow\)x=3
\(B=-2x^2+8x-15\)
\(-2B=4x^2-16x+30\)
\(-2B=\left[\left(2x\right)^2-2.2x.4+4^2\right]+14\)
\(-2B=\left(2x-4\right)^2+14\)
\(\Rightarrow B=-\frac{\left(2x-4\right)^2}{2}-7\)
Ta có: \(-\frac{\left(2x-4\right)^2}{2}\le0\forall x\)
Đến đây b làm tương tự như trên nhé.
Chúc b học tốt
a) \(A=-x^2+6x-15\)
\(-A=x^2-6x+15\)
\(-A=\left(x^2-6x+9\right)+6\)
\(-A=\left(x-3\right)^2+6\)
Mà \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-A\ge6\)
\(\Leftrightarrow A\le-6\)
Dấu "=" xảy ra khi :
\(x-3=0\Leftrightarrow x=3\)
Vậy \(A_{Max}=-6\Leftrightarrow x=3\)
b) \(B=-2x^2+6x-15\)
\(-2B=4x^2-12x+30\)
\(-2B=\left(4x^2-12x+9\right)+21\)
\(-2B=\left(2x-3\right)^2+21\)
Mà \(\left(2x-3\right)^2\ge0\forall x\)
\(\Rightarrow-2B\ge21\)
\(\Leftrightarrow B\le-\frac{21}{2}\)
Dấu "=" xảy ra khi :
\(2x-3=0\Leftrightarrow x=\frac{3}{2}\)
Vậy \(B_{Max}=-\frac{21}{2}\Leftrightarrow x=\frac{3}{2}\)
