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thaonguyenkmhd


*Ta có: $x<y \rightarrow \dfrac{a}{m} < \dfrac{b}{m} \rightarrow \dfrac{a+b}{m} < \dfrac{2b}{m} \rightarrow \dfrac{a+b}{2m} <\dfrac{2b}{2m} \rightarrow \dfrac{a+b}{2m} < \dfrac{b}{m} \rightarrow z < y (1)$

*Ta có: $x<y \rightarrow \dfrac{a}{m} < \dfrac{b}{m} \rightarrow \dfrac{2a}{m} < \dfrac{a+b}{m} \rightarrow \dfrac{2a}{2m} < \dfrac{a+b}{2m} \rightarrow \dfrac{a}{m}< \dfrac{a+b}{2m} \rightarrow x < z (2)$

Từ (1) và (2) \Rightarrow x<z<y (đpcm)
 
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