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[TEX] B= \frac{5n+11}{3n+2}[/TEX]
tìm max B với n thuộc Z
làm thử đi
[TEX] B= \frac{5n+11}{3n+2}[/TEX]
\Rightarrow [TEX] B= \frac{3(5n+11)}{3(3n+2)}[/TEX]
\Rightarrow [TEX] B= \frac{5(3n+11)}{3(3n+2)}[/TEX]
\Rightarrow [TEX] B= \frac{5(3n+2)+13}{3(3n+2)}[/TEX]
\Rightarrow [TEX]B[/TEX]= [TEX]\frac{5}{3}[/TEX] + [TEX]\frac{13}{3(3n+2)}[/TEX]
Để B có GTLN thì [TEX]M = \frac{13}{3(3n+2)}[/TEX] có GTLN
(*)Nếu 3(3n+2)< 0 thì M < 0 (1)
(*) Nếu 3(3n+2) > 0 \Rightarrow 3(3n+2) \geq 3 ( do n nguyên )
\Rightarrow [TEX]\frac{1}{3(3n+2)}[/TEX] \leq [TEX]\frac{1}{3}[/TEX]
\Rightarrow [TEX]\frac{13}{3(3n+2)}[/TEX] \leq [TEX]\frac{13}{3}[/TEX]
hay M \leq [TEX]\frac{13}{3}[/TEX] (2)
\Rightarrow max M = [TEX]\frac{13}{3}[/TEX] khi 3(3n+2) = 3 \Rightarrow 3n+2 = 1 \Rightarrow 3n = -1 \Rightarrow n = [TEX]\frac{-1}{3}[/TEX]
So sánh (1) và (2) \Rightarrow max M = [TEX]\frac{13}{3}[/TEX] khi n = [TEX]\frac{-1}{3}[/TEX]
\Rightarrow max B = [TEX]\frac{5}{3}+\frac{13}{3}[/TEX] = [TEX]\frac{18}{3}[/TEX] = 6 khi = [TEX]\frac{-1}{3}[/TEX]
Vậy max B = 6 khi = [TEX]\frac{-1}{3}[/TEX]
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