Question

[nguyễn minh chuyên]

cho b>0. So sánh \(\frac{a}{b}\)và \(\frac{a+1}{b+1}\)

Answer 1

[Thanh Tùng DZ]

Xét 3 trường hợp :

+) Nếu b > a thì \(\frac{a}{b}=\frac{b-m}{b}=\frac{b}{b}-\frac{m}{b}=1-\frac{m}{b}\)

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

Vì \(\frac{m}{b}>\frac{m}{b+1}\)nên \(1-\frac{m}{b}< 1-\frac{m}{b+1}\)hay \(\frac{a}{b}< \frac{a+1}{b+1}\)

+) Nếu a = b thì \(\frac{a}{b}=1\)

\(\frac{a+1}{b+1}=1\)nên\(\frac{a}{b}=\frac{a+1}{b+1}\)

+) Nếu a > b thì \(\frac{a}{b}=\frac{b+m}{b}=\frac{b}{b}+\frac{m}{b}=1+\frac{m}{b}\)

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

Vì \(\frac{m}{b}>\frac{m}{b+1}\)nên \(1+\frac{m}{b}>1+\frac{m}{b+1}\)hay \(\frac{a}{b}>\frac{a+1}{b+1}\)

Answer 2

[Kurosaki Akatsu]

Ta có :

\(\frac{a}{b}=\frac{a\left(b+1\right)}{b\left(b+1\right)}=\frac{ab+a}{b^2+b}\)

\(\frac{a+1}{b+1}=\frac{b\left(a+1\right)}{b\left(b+1\right)}=\frac{ab+b}{b^2+b}\)

Từ 2 ý trên , ta xét từng trường hợp sau :

a < b thì \(\frac{a}{b}< \frac{a+1}{b+1}\)

a > b thì \(\frac{a}{b}>\frac{a+1}{b+1}\)

a = b thì \(\frac{a}{b}=\frac{a+1}{b+1}\)