[Toán 8] Tìm GTNN và GTLN

N

nguyenbahiep1

[laTEX]A = \frac{x^2+x+1}{x^2-x+1} \\ \\ x^2-x +1 = (x-\frac{1}{2})^2 + \frac{3}{4} > 0 \\ \\ A = \frac{1}{3} + \frac{2(x+1)^2}{3(x^2-x+1)} \geq \frac{1}{3} \Rightarrow x = - 1 \\ \\ A = 3 - \frac{2(x-1)^2}{x^2-x+1} \leq 3 \Rightarrow x = 1 \\ \\ GTNN_A = \frac{1}{3} , GTLN_A = 3[/laTEX]
 
D

dien0709

2)[TEX]B=2+\frac{2x}{x^2+1}=2+1+\frac{2x}{x^2+1}-1[/TEX]=[TEX]\1+1+\frac{2x}{x^2+1}=3-\frac{(x-1)^2}{x^2+1}=1+\frac{(x+1)^2}{x^2+1}[/TEX]
Vậy [TEX]B\leq3[/TEX]=>maxB=3,[TEX]B\geq1[/TEX]=>minB=1
 
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