Đáp án:
Giải thích các bước giải:
$\text{Bài 1:}$
$\text{a, A= $x^{6}$ + 5}$
$\text{Vì $x^{6}$ ≥ 0 ∀ x}$
$\text{⇒$x^{6}$ + 5 ≥ 5 ∀ x}$
$\text{Dấu = xảy ra ⇔ x=0}$
$\text{Vậy $Min_{A}$ = 5 ⇔ x=0}$
$\text{b, B= (x-7)²-3}$
$\text{Vì (x-7)² ≥ 0 ∀ x}$
$\text{⇒(x-7)² - 3≥ -3 ∀ x}$
$\text{Dấu = xảy ra ⇔ x-7=0 ⇔ x=7}$
$\text{Vậy $Min_{B}$ = -3 ⇔ x=7}$
$\text{c, C= x²-4x+8}$
$\text{C=x²-4x+4+4}$
$\text{C=(x-2)²+4}$
$\text{Vì (x-2)² ≥ 0 ∀ x}$
$\text{⇒(x-2)² +4≥ 4 ∀ x}$
$\text{Dấu = xảy ra ⇔ x-2=0 ⇔ x=2}$
$\text{Vậy $Min_{C}$ = 4 ⇔ x=2}$
$\text{d, D= 5x²-10x+1}$
$\text{D= x²+4x²-2x-8x+1+4-4}$
$\text{D=(x²-2x+1)+(4x²-8x+4)-4}$
$\text{D=(x-1)²+(2x-2)²-4}$
$\text{Vì (x-1)² ≥ 0 ∀ x, (2x-2)² ≥ 0 ∀ x}$
$\text{⇒(x-2)²+ (2x-2)²-4≥ -4 ∀ x}$
$\text{⇒D ≥-3}$
$\text{Dấu = xảy ra ⇔ 2x-2=0 ⇔ x=1}$
$\text{Vậy $Min_{D}$ = -3 ⇔ x=1}$
$\text{e, E=x²-2x+1+y²-4y+5}$
$\text{E=(x²-2x+1)+(y²-4y+4)+1}$
$\text{E=(x-1)²+(y-2)²+1}$
$\text{Vì (x-1)² ≥ 0 ∀ x, (y-2)² ≥ 0 ∀ x}$
$\text{⇒(x-1)²+ (y-2)²+1≥ 1 ∀ x}$
$\text{Dấu = xảy ra ⇔ x-1=0 ⇔ x=1 và y-2=0 ⇔ y=2}$
$\text{Vậy $Min_{E}$ = 1 ⇔ x=1 và y=2}$
$\text{Bài 2:}$
$\text{a,A=-x²-2x+5}$
$\text{A=-(x²+2x-5)}$
$\text{A=-(x²+2x+1)-6}$
$\text{A=-(x+1)²+6}$
$\text{Vì -(x-1)² ≤ 0 ∀ x}$
$\text{⇒-(x+1)² + 6 ≤ 6 ∀ x}$
$\text{Dấu = xảy ra ⇔ x+1=0 ⇔ x=-1}$
$\text{Vậy $Max_{A}$ = 6 ⇔ x=-1}$
$\text{b,B=-3x²-6x-2}$
$\text{B=-3(x²+2x+$\frac{2}{3}$ }$
$\text{B= -3(x²+2x+1-$\frac{1}{3}$ }$
$\text{B=-3(x+1)²+1}$
$\text{Vì -3(x-1)² ≤ 0 ∀ x}$
$\text{⇒-3(x+1)² + 1 ≤ 1 ∀ x}$
$\text{Dấu = xảy ra ⇔ x+1=0 ⇔ x=-1}$
$\text{Vậy $Max_{B}$ = 1 ⇔ x=-1}$
$\text{c, C= -2x(x-5)+3}$
$\text{C=-2x²+10x+3}$
$\text{C= -2(x²-5x- $\frac{3}{2}$ )}$
$\text{C=-2(x²-5x+$\frac{25}{4}$ - $\frac{31}{4}$ )}$
$\text{C=-2[(x-$\frac{5}{2}$ )²-$\frac{31}{4}$ ]}$
$\text{C= -2(x-$\frac{5}{2}$ )² + $\frac{31}{2}$ }$
$\text{Vì -2(x-$\frac{5}{2}$ )² ≤ 0 ∀ x}$
$\text{⇒-2(x-$\frac{5}{2}$ )² + $\frac{31}{2}$ ≤ $\frac{31}{2}$ ∀ x}$
$\text{Dấu = xảy ra ⇔ x-$\frac{5}{2}$=0 ⇔ x=$\frac{5}{2}$ }$
$\text{ Vậy $Max_{C}$ = $\frac{31}{2}$ ⇔ x=$\frac{5}{2}$ }$
