So sánh \(\frac{a}{b}\) (b > 0) và \(\frac{a+n}{b+n}\), (\(n \in Z\))
hay a(b + n) < b.(a + n)
=> \(\frac{a}{b}\) < \(\frac{a+n}{b+n}\)
hay a(b + n) > b(a + n) => \(\frac{a}{b}\) > \(\frac{a+n}{b+n}\)
TH3: Nếu a = b => \(\frac{a}{b}\) = \(\frac{a+n}{b+n}\) (=1)
-- Mod Toán 7
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