Bài $5$.
$a$) `A = 3|2x-1| - 1`
Vì : $|2x-1| ≥ 0 ∀ x$
$⇒$ $A = 3|2x-1| - 1 ≥ 3.0 - 1 = -1$. Dấu " $=$ " khi:
$2x-1 =0 ⇔ x = \dfrac{1}{2}$
Vậy $A_{min} = -1$ khi $x = \dfrac{1}{2}$
`B = |x+1| + 2(6,3 - 3y)^2 + 3`
Vì : `|x+1|;|(6,3-3y)^2 ≥ 0 ∀ x;y`
`B = |x+1| + 2(6,3 - 3y)^2 + 3 ≥ 0 + 0 + 3 = 3` . Dấu " $=$ " khi:
$\left\{\begin{matrix}x = -1 & \\ y = 2,1& \end{matrix}\right.$
Vậy `B_{min} = 3` khi `(x;y)=(-1;2,1)`
$b$) `C = 8 - 6|x-7| - (y^2 - 16)^2`
Vì : $|x-7| ; (y^2 - 16)^2 ≥ 0 ∀ x;y$
$⇒ C = 8 - 6|x-7| - (y^2 - 16)^2 ≤ 8 - 0 - 0 = 8$. Dấu " $=$ " khi:
$\left\{\begin{matrix}x =7 & \\ y = ±4& \end{matrix}\right.$
Vậy `C_{max} = 8` khi `(x;y)=(7;4);(7;-4)`
$c$) `x^2 + (y-1)^2 = 0`
Vì : `x^2;(y-1)^2 ≥ 0 ∀ x;y`
$⇒$ $x^2 = (y-1)^2 = 0$
`⇒` $\left\{\begin{matrix}x = 0 & \\ y = 1& \end{matrix}\right.$
Vậy `(x;y)=(0;1)`
