[Toán 7] Chứng minh

H

harrypham

Ta có $x^m= \underbrace{x.x....x}_{m \; \text{số}}, \; x^n= \underbrace{x.x....x}_{n \; \text{số}}$ nên $$\dfrac{x^m}{x^n}= \dfrac{ \underbrace{x.x....x}_{m \; \text{số}}}{ \underbrace{x.x....x}_{n \; \text{số}}} = \underbrace{x.x....x}_{m-n \; \text{số}}=x^{m-n}$$
 
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