Question

[Nguyễn Hồng Nhung]

Bài 3: Tìm GTNN
A = 4x² - 2x + 1
B = x⁴ - 4x²
C = x² - 2x + 3 với x≥2
D = (x² - x + 1) (x² - x - 1)
E = (x - 2)² + (x - 4)²
F = x⁴ - 6x³ + 10x² - 6x + 9
G = x⁴ + 4x³ + 10x² + 12x + 11
H = x⁴ - 6x³ + x² + 24x +18
I = x⁴ + 4x³ - x² - 14x + 6
K = x⁴ + 2x³ - 10x² - 16x + 45
<Giúp với>

Answer 1

[Akai Haruma]

i)

$I=x^4+4x^3-x^2-14x+6$

$=(x^4+4x^4+4x^2)-5x^2-14x+6$

$=(x^2+2x)^2-6(x^2+2x)+9+x^2-2x-3$

$=(x^2+2x-3)^2+(x^2-2x+1)-4$

$=(x-1)^2(x+3)^2+(x-1)^2-4$

$=(x-1)^2[(x+3)^2+1]-4\geq -4$

Vậy $I_{\min}=-4$ khi $(x-1)^2[(x+3)^2+1]=0\Leftrightarrow x=1$

k)

$K=x^4+2x^3-10x^2-16x+45$

$=(x^4+2x^3+x^2)-11x^2-16x+45$

$=(x^2+x)^2-12(x^2+x)+x^2-4x+45$

$=(x^2+x)^2-12(x^2+x)+36+(x^2-4x+4)+5$

$=(x^2+x-6)^2+(x-2)^2+5$

$=[(x-2)(x+3)]^2+(x-2)^2+5$

$=(x-2)^2[(x+3)^2+1]+5\geq 5$

Vậy $K_{\min}=5$ khi $(x-2)^2[(x+3)^2+1]=0\Leftrightarrow x=2$

Answer 2

[Akai Haruma]

g)

$G=x^4+4x^3+10x^2+12x+11$

$=(x^4+4x^3+4x^2)+6x^2+12x+11$

$=(x^2+2x)^2+6(x^2+2x)+11$

Đặt $x^2+2x=t$. Khi đó $t=x^2+2x=(x+1)^2-1\geq -1\Rightarrow t+1\geq 0$

$\Rightarrow G=t^2+6t+11=(t+1)^2+4(t+1)+7\geq 7$

Vậy $G_{\min}=7$ khi $t=-1\Leftrightarrow (x+1)^2=0\Leftrightarrow x=-1$

h)

$H=x^4-6x^3+x^2+24x+18$

$=(x^4-6x^3+9x^2)-8x^2+24x+18$

$=(x^2-3x)^2-8(x^2-3x)+18$

$=(x^2-3x)^2-8(x^2-3x)+16+2$

$=(x^2-3x-4)^2+2\geq 2$

Vậy $H_{\min}=2$ khi $x^2-3x-4=0\Leftrightarrow x=4$ hoặc $x=-1$