Đáp án + Giải thích các bước giải:
Bài 2: $B(x)=2x^2-2(x-5)=2(x^2-x+5)=2\left(x^2-x+\dfrac14+\dfrac{19}4\right)\\=2\left[\left(x-\dfrac14\right)^2+\dfrac{19}4\right]=2\left(x-\dfrac14\right)^2+8$
Vì $2\left(x-\dfrac14\right)^2≥0\ ∀ x\in \mathbb{R}$
$⇒ 2\left(x-\dfrac14\right)^2+8 ≥8\ ∀ x\in \mathbb{R}$
Vậy $\min B(x)=8$ khi $x-\dfrac14=0 ⇔ x=\dfrac14$
Bài 3:$C(y)=(y+2)^2+(y-5)^2=y^2+4y+4+y^2-10y+25\\=2y^2-6y+29=2\left(y^2-3y+\dfrac{29}2\right)=2\left(y^2-3y+\dfrac{9}4+\dfrac{49}4\right)\\=2\left[\left(y-\dfrac32\right)^2+\dfrac{49}4\right]=2\left(y-\dfrac32\right)^2+\dfrac{49}2$
Vì $2\left(y-\dfrac32\right)^2≥0\ ∀ x\in \mathbb{R}$
$⇒ 2\left(y-\dfrac32\right)^2+\dfrac{49}2≥\dfrac{49}2\ ∀ x\in \mathbb{R}$
Vậy $\min C(y)=\dfrac{49}2$ khi $y-\dfrac32=0 ⇔ y=\dfrac32$
Bài 4: $A(x;y)=x^2+2x+9y^2-6y+2018\\ =x^2+2x+1+9y^2-6y+1+2016\\ =(x+1)^2+(3y+1)^2+2016$
Vì $(x+1)^2+(3y+1)^2≥0\ ∀ x\in \mathbb{R}$
$⇒ (x+1)^2+(3y+1)^2+2016≥2016\ ∀ x\in \mathbb{R}$
Vậy $\min A(x;y)=2016$ khi $\begin{cases}x+1=0\\3y+1=0\end{cases}⇔\begin{cases}x=-1\\y=\dfrac{-1}3\end{cases}$
