\(H=3-16x^2+32x\)
\(H=-\left(16x^2-32x-3\right)\)
\(H=-\left[\left(4x\right)^2-2\cdot4x\cdot4+16-19\right]\)
\(H=-\left[\left(4x-4\right)^2-19\right]\)
\(H=19-\left(4x-4\right)^2\le19\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
Ta có: \(H=3-16x^2+32x\)
\(=67-64-16x^2+32x\)
\(=67-\left(16x^2-32x+64\right)\)
\(=67-\left(4x-8\right)^2\)
Nhân xét: \(\left(4x-8\right)^2\ge0\)
\(\Rightarrow H\le67\)
H đạt GTLN ⇔ \(\left(4x-8\right)^2=0\)
\(\Leftrightarrow4x-8=0\Leftrightarrow x=2\)
Vậy \(H_{MAX}=67\Leftrightarrow x=2\)
Chúc bạn học tốt@@
H = 3 - 16x2 + 32x
= -(16x2 - 32x - 3)
= -[(4x)2 - 2.4x.4 + 42 - 19)]
= -[(4x+16)2 - 19]
= -(4x + 16)2 +19
Vì -(4x + 16)2 ≤ 0 ∀x ⇒ -(4x + 16)2 +19 ≤ 19 ∀x
⇒ H ≤ 19 ∀x
Dấu "=" xảy ra ⇔ -(4x + 16)2 = 0 ⇔ 4x + 16 = 0 ⇔ x = -4
Vậy maxH = 19 ⇔ x = -4