Tìm GTLN

B

beconvaolop

[TEX]\frac{4x+1}{x^2+5}= \frac{x^2+5-x^2+4x-4}{x^2+5}= 1- \frac{(x-2)^2}{x^2+5}[/TEX]
BT max\Leftrightarrow x-2=0 \Leftrightarrow x=2
 
N

niemkieuloveahbu

[TEX]A=\frac{4x+1}{x^2+5}\\ \Leftrightarrow Ax^2-4x+5A-1=0\\ A=0\Rightarrow x=\frac{-1}{4}\\ A\neq 0,pt\ co\ nghiem: \\ \Leftrightarrow \Delta' \geq 0\Leftrightarrow 4-5A^2+1\geq 0\Rightarrow -1 \leq A\leq1[/TEX]

Thay a => x
 
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