Question

[Hikaru Sumire]

cho a,b,c thuộc n chứng minh : nếu a>b thì a/b>a+c/b+c

Answer 1

[.]

Vì \(a>b\) nên \(a=b+m\)   \(\left(m\inℕ^∗\right)\)

Ta có : \(\frac{a}{b}=\frac{b+m}{b}=1+\frac{m}{b}\)

         \(\frac{a+c}{b+c}=\frac{b+m+c}{b+c}=1+\frac{m}{b+c}\)

Mà \(\frac{m}{b}>\frac{m}{b+c}\) nên \(1+\frac{m}{b}>1+\frac{m}{b+c}\)

hay \(\frac{a}{b}>\frac{a+c}{b+c}\)   (đpcm)

Answer 2

[๖²⁴ʱ๖ۣۜTɦủү❄吻༉]

Theo cj nghĩ : 

\(a>b\Rightarrow a-b>0\left(a;b\inℕ^∗\right)\)

Mà : \(\frac{a}{b}-\frac{a+c}{b+c}=\frac{a\left(b+c\right)}{b\left(b+c\right)}-\frac{b\left(a+c\right)}{b\left(b+c\right)}=\frac{a\left(b+c\right)-b\left(a+c\right)}{b\left(b+c\right)}=\frac{c\left(a-b\right)}{b\left(b+c\right)}>0\)

Do đó : \(\frac{a}{b}>\frac{a+c}{b+c}\left(đpcm\right)\)