Question

cho a/b < 1. cmr a/b < a+c/b+c       (a,b,c thuộc th Z, b,c >0)

Answer 1

Vì \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{b}{b}\Rightarrow a< b\)  (vì b >0)

Có : \(\frac{a}{b}=\frac{a\left(b+c\right)}{b\left(b+c\right)}=\frac{ab+ac}{b\left(b+c\right)}\)

        \(\frac{a+c}{b+c}=\frac{b\left(a+c\right)}{b\left(b+c\right)}=\frac{ab+bc}{b\left(b+c\right)}\)

Vì b,c > 0 => b + c > 0 => b(b+c) > 0

Vì a < b , c>0 => ac < bc => \(\frac{ab+ac}{b\left(b+c\right)}< \frac{ab+bc}{b\left(b+c\right)}\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)

Answer 2

Ta có:

(a + c) / (b + c) = (a + b + c) / (b + c) - b/(b + c) = a/(b + c) + 1 - b/(b + c) (1)

Mà a/b < 1 nên a < b (2)

Từ (1),(2) suy ra:

a/(b + c) - b/(b + c) + 1 < 1

Vậy nên ta có a/b < (a + c) /(b+c)