A = -a2 + 3a + 4
A = -( a2 - 3a + 9/4 ) + 25/4
A = -( a - 3/2 )2 + 25/4
-( a - 3/2 )2 ≤ 0 ∀ x => -( a - 3/2 )2 + 25/4 ≤ 25/4
Đẳng thức xảy ra <=> a - 3/2 = 0 => a = 3/2
=> MaxA = 25/4 <=> a = 3/2
\(A=-a^2+3a+4\)
\(\Rightarrow A=-a^2+3a-\frac{9}{4}+\frac{25}{4}\)
\(\Rightarrow A=-\left(a-\frac{3}{2}\right)^2+\frac{25}{4}\)
Vì \(\left(a-\frac{3}{2}\right)^2\ge0\forall a\)\(\Rightarrow-\left(a-\frac{3}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(a-\frac{3}{2}\right)^2=0\Leftrightarrow a-\frac{3}{2}=0\Leftrightarrow a=\frac{3}{2}\)
Vậy maxA = 25/4 <=> a = 3/2
\(-a^2+3a+4\)
=\(-a^2+3a-\frac{9}{4}+\frac{25}{4}\)
=\(-\left(a-\frac{3}{2}\right)^2+\frac{25}{4}\)
Ta có : \(\left(a-\frac{3}{2}\right)^2\ge0\forall a\)
\(-\left(a-\frac{3}{2}\right)^2\le0\)
\(-\left(a-\frac{3}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Dấu = xảy ra :
\(\Leftrightarrow\left(a-\frac{3}{2}\right)^2=0\)
\(a-\frac{3}{2}=0\)
\(a=\frac{3}{2}\)
Vậy A đạt giá trị lớn nhất = 25/4 khi và chỉ khi a =3/2
\(A=-a^2+3a+4\)
\(A=-\left(a^2-3a-4\right)\)
\(A=-\left(a^2-3a+\frac{9}{4}\right)^2+\frac{25}{4}\)
\(A=-\left(a-\frac{3}{2}\right)+\frac{25}{4}\le\frac{25}{4}\)
Max A = 25/4 <=> a = 3/2
\(A=-a^2+3a+4\)
\(\Rightarrow A=-a^2+3a-\frac{9}{4}+\frac{25}{4}\)
\(\Rightarrow A=-\left(a-\frac{3}{2}\right)^2+\frac{25}{4}\)
Ta có : \(\left(a-\frac{3}{2}\right)^2\ge0\forall a\)
\(-\left(a-\frac{3}{2}\right)^2\le0\)
\(-\left(a-\frac{3}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Dấu '=' xảy ra : \(\Leftrightarrow-\left(a-\frac{3}{2}\right)^2=0\Leftrightarrow a-\frac{3}{2}=0\Leftrightarrow a=\frac{3}{2}\)
Vậy maxA \(=\frac{25}{4}\)\(\Rightarrow a=\frac{3}{2}\)
