[Toán 6] So sánh

T

tanngoclai

Ta có :
[TEX] A = \frac{10}{a^m} + \frac{10}{a^n} = \frac{10}{a^m} + \frac{9}{a^n} + \frac{1}{a^n} [/TEX]

[TEX] B = \frac{11}{a^m} + \frac{9}{a^n} = \frac{10}{a^m} + \frac{9}{a^n} + \frac{1}{a^m} [/TEX]

- Với : [TEX] m=n \Rightarrow a^n = a^m \Rightarrow \frac{1}{a^m} = \frac{1}{a^n} \Rightarrow \frac{10}{a^m} + \frac{9}{a^n} + \frac{1}{a^m} = \frac{10}{a^m} + \frac{9}{a^n} + \frac{1}{a^m} \Rightarrow A = B[/TEX]

- Với : [TEX] m < n \Rightarrow a^m < a^n \Rightarrow \frac{1}{a^n} < \frac{1}{a^m} \Rightarrow \frac{10}{a^m} + \frac{9}{a^n} + \frac{1}{a^n} < \frac{10}{a^m} + \frac{9}{a^n} + \frac{1}{a^m} \Rightarrow A < B[/TEX]

- Với : [TEX] m > n \Rightarrow a^m > a^n \Rightarrow \frac{1}{a^n} > \frac{1}{a^m} \Rightarrow \frac{10}{a^m} + \frac{9}{a^n} + \frac{1}{a^n} > \frac{10}{a^m} + \frac{9}{a^n} + \frac{1}{a^m} \Rightarrow A > B[/TEX]
 
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