Toán12 so sánh lũy thừa

T

trang020192

tham khảo thử nhá :D
Xét[TEX] f(x)=\frac{x}{lnx} --> f'(x)>0, \forall x>e[/TEX]
=> [TEX]\frac{x}{lnx}>\frac{x+1}{ln(x+1)}[/TEX]
=> [TEX]xln(x+1)>(x+1)lnx[/TEX]
=>[TEX] (x+1)^x > x^(x+1)....[/TEX]
 
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