Toán Lớp 8: GTLN:
6x – x^2 + 3
12x – 8y – 4x^2 – y^2 +1
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: GTLN:
6x – x^2 + 3
12x – 8y – 4x^2 – y^2 +1
Comments ( 2 )
$\text{A=-($x^2$ – 6x – 3)}$
$\text{A=-($x^2$ – 2.3.x + $3^2$ – $3^2$ – 3)}$
$\text{A=-($(x-3)^2$ – 12)}$
$\text{A=-$(x-3)^2$ + 12 $\le$ 12}$
$\text{Vậy GTLN A=12 ($\forall$x) khi x-3=0}$
$\text{=>x=3}$
$\text{Gọi B= 12x – 8y – $4x^2$ – $y^2$ +1}$
$\text{B= 12x – 8y – $4x^2$ – $y^2$ + 1}$
$\text{B=-[($4x^2$ – 12x)+($y^2$ + 8y) – 1]}$
$\text{B=-[(($2x)^2$ – 2.2x.3 + $3^2$)+($y^2$ + 2.y.4 + $4^2$) – $3^2$-$4^2$-1]}$
$\text{B=-[$(2x-3)^2$ + $(y+4)^2$ – 26]}$
$\text{B=-$(2x-3)^2$ – $(y+4)^2$ + 26 $\le$ 26 }$
$\text{Vì -$(2x-3)^2$ $\le$ 0 ($\forall$x)}$
$\text{Vì -$(y+4)^2$ $\le$ 0 ($\forall$y)}$
$\text{Vậy để GTLN B=26 khi:}$
$\text{$\left[\begin{matrix} 2x – 3 = 0\\ y + 4 = 0\end{matrix}\right.$}$
$\text{$\left[\begin{matrix} x = \dfrac{3}{2}\\ y = -4\end{matrix}\right.$}$