Question

[mai thị hà vi]

cho a,b,c là số nguyên và b\(\ne\)0

so sánh \(\frac{a}{b}\)và \(\frac{a+n}{b+n}\)

Answer 1

[I don't need you]

ta có: \(a.\left(b+n\right)=ab+an;b.\left(a+n\right)=ba+bn\)

nếu a < b

=> ab + an < ba + bn

=> a.(b+n) < b.(a+n)

\(\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\)

nếu a = b

...

---> a/b = a+n/b+n

nếu a > b

...

----> a/b > a+n/b+n

Answer 2

[kudo shinichi]

Theo mk thì \(a,b,n\in N\)

Xét hiệu:

\(\frac{a}{b}-\frac{a+n}{b+n}=\frac{a.\left(b+n\right)-\left(a+n\right).b}{b.\left(b+n\right)}=\frac{an-bn}{b\left(b+n\right)}=\frac{n\left(a-b\right)}{b.\left(b+n\right)}\)

Với \(a=b\Rightarrow a-b=0\Rightarrow\frac{n.\left(a-b\right)}{b.\left(b+n\right)}=0\Rightarrow\frac{a}{b}=\frac{a+n}{b+n}\)

Với \(a>b\Rightarrow a-b>0\Rightarrow\frac{n.\left(a-b\right)}{b.\left(b+n\right)}>0\Rightarrow\frac{a}{b}>\frac{a+n}{b+n}\)

Với \(a< b\Rightarrow a-b< 0\Rightarrow\frac{n.\left(a-b\right)}{b.\left(b+n\right)}< 0\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\)

Vậy \(\frac{a}{b}=\frac{a+n}{b+n}\Leftrightarrow a=b\)

      \(\frac{a}{b}>\frac{a+n}{b+n}\Leftrightarrow a>b\)

      \(\frac{a}{b}< \frac{a+n}{b+n}\Leftrightarrow a< b\)

Tham khảo nhé~