giup minh bai nay voi,chieu nay kiem tra dang nay ma gio khong biet lam...

D

duynhan1

cobala8294 said:
tim a,b de:
A = lim(x tien den duong vo cung) (\sqrt{x2+x+1}-ax- b)=0
Bấm để xem đầy đủ nội dung ...

[TEX]A = \lim_{x \to + \infty} (\frac{(1-a^2) x^2 + x +1}{\sqrt{x^2+x+1} + ax} - b \\ = \lim_{x \to + \infty} \frac{(1-a)x + 1 + \frac{1}{x} }{ \sqrt{1+ \frac{1}{x} + \frac{1}{x^2} } + a} - b [/TEX]

[TEX] * a \not=1 \Rightarrow A = \infty [/TEX]

[TEX]* a = 1 \Rightarrow A =\lim_{x \to \infty} \frac{ 1 + \frac{1}{x} }{ \sqrt{1+ \frac{1}{x} + \frac{1}{x^2} } + 1} - b = \frac12 - b [/TEX]

[TEX]A = 0 \Leftrightarrow b = \frac12[/TEX]

[TEX] KL: \left{ a= 1 \\ b = \frac12[/TEX]
 
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