Ta có: x < y \(\Rightarrow\) \(\dfrac{a}{m}\)<\(\dfrac{b}{m}\) \(\Rightarrow\) am < bm (m > 0) \(\Rightarrow\) am + am < bm + am \(\Rightarrow\) 2am < m (b + a) \(\Rightarrow\) \(\dfrac{2a}{m}< \dfrac{a+b}{m}\) \(\Rightarrow\) \(\dfrac{a}{m}< \dfrac{a+b}{m}\). Vậy x < r ( 1 )
T. Tự, ta có: x < y \(\Rightarrow\) \(\dfrac{a}{m}< \dfrac{b}{m}\)\(\Rightarrow\) am < bm (m > 0) \(\Rightarrow\) am + bm < bm + bm \(\Rightarrow\) m ( a + b ) < 2bm \(\Rightarrow\) \(\dfrac{2\left(a+b\right)}{m}< \dfrac{b}{m}\) \(\Rightarrow\dfrac{a+b}{m}< \dfrac{b}{m}\). Vậy r < y (2)