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Bài1. Tìm giá trị lớn nhất , giá trị nhỏ nhất có thể
a. x^2-5x+8
b. 4-2x-x^2
c. 11-7x-x^2
d. 9+3x-x^2
f.4x^2+8x+11
g.9x^2+x+1
h.16x^2-3x+1:confused::confused::confused::confused::confused::confused:

a,
[TEX] x^2-5x+8=(x^2-5x+\frac{25}{4})+\frac{7}{4}=(x-\frac{5}{2})^2 +\frac{7}{4}\geq \frac{7}{4}[/TEX]

Dấu = xảy ra khi[TEX] x=\frac{5}{2}[/TEX]

b,
[TEX] 4-2x-x^2= -(x^2+2x+1)+5 = -(x+1)^2+5 \leq 5[/TEX]

Dấu = xảy ra khi[TEX] x=-1[/TEX]
c,
[TEX] 11-7x-x^2=-(x^2+7x+\frac{49}{4})+\frac{93}{4} =-(x+\frac{7}{2})^2+\frac{93}{4}\leq \frac{93}{4}[/TEX]

Dấu = xảy ra khi [TEX]x=-\frac{7}{2}[/TEX]
d,

[TEX]9+3x-x^2=-(x^2-3x+\frac{9}{4})+\frac{45}{4}=-(x-\frac{3}{2})^2+\frac{45}{4}\leq 45/4[/TEX]

Dấu = xảy ra khi [TEX]x=\frac{3}{2}[/TEX]
f,
[TEX]4x^2+8x+11= 4(x^2+2x+1)+7=4(x+1)^2+7 \geq 7[/TEX]

Dấu = xảy ra khi [TEX]x=-1[/TEX]
g,
[TEX] 9x^2+x+1=(9x^2+2.3x.\frac{1}{6}+\frac{1}{36})+\frac{35}{36}=(3x+\frac{1}{6})^2+\frac{35}{36} \geq \frac{35}{36}[/TEX]

Dấu = xảy ra khi.........

h,
[TEX] 16x^2-3x+1= (16x^2-2.4x.\frac{3}{8}+\frac{9}{64})+\frac{55}{64}=(4x-\frac{3}{8})^2+\frac{55}{64} \geq \frac{55}{64}[/TEX]

Dấu = xảy ra khi.....
 
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