bài toán về lũy thừa..

N

nguyenbahiep1

([TEX]sqrt{10}+3[/TEX])^[TEX]\frac{x-3}{x-1}[/TEX] = ([TEX]sqrt{10}-3[/TEX])^[TEX]\frac{x+1}{x+3}[/TEX]


ta có

[TEX](\sqrt{10}+ 3)^{-1} = (\sqrt{10}- 3)[/TEX]

từ đây ta có

[TEX](\sqrt{10}+ 3)^{\frac{x-3}{x-1}} = (\sqrt{10}+ 3)^{-\frac{x+1}{x+3}} \\ \frac{x-3}{x-1} = -\frac{x+1}{x+3} \\ x^2-9 = -(x^2-1) \Rightarrow x^2 = 5 \Rightarrow x = \pm \sqrt{5}[/TEX]
 
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