Question

Cho a, b, n \(\in\)N*. Hãy so sánh \(\frac{a+n}{b+n}\)và \(\frac{a}{b}\)

Answer 1

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

\(\frac{a}{b}=\frac{a\left(b+n\right)}{b\left(b+n\right)}=\frac{ab+an}{b^2+bn}\)

2 phân thức cùng mẫu, ta so sánh tử số 

+) TH1 : a > b => an > bn

=> \(\frac{a}{b}>\frac{a+n}{b+n}\)

+) TH2 : a < b => an < bn

=> \(\frac{a}{b}< \frac{a+n}{b+n}\)

+) TH3 : a = b => an = bn

=> \(\frac{a}{b}=\frac{a+n}{b+n}\)

Answer 2

Ta co: (a+n).b=a.b+n.b

(b+n).a=b.a+n.a

Xet tuong hop:

Th1: a>b

Voi a>b thi a.b+n.b<b.a+n.a 

​​a+n/b+n<a/b

Th2:b>a

Voi b>a thi a.b+b.a>b.a+n.a

a+n/b+n>a/b

Answer 3

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

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

Nếu \(a\ge b\Rightarrow b-a\le0\Rightarrow\frac{a+n}{b+n}-\frac{a}{b}\le0\Rightarrow\frac{a+n}{b+n}\le\frac{a}{b}\)

Nếu \(a\le b\Rightarrow b-a\ge0\Rightarrow\frac{a+n}{b+n}-\frac{a}{b}\ge0\Rightarrow\frac{a+n}{b+n}\ge\frac{a}{b}\)

Vậy xảy ra 2 trường hợp:

\(\frac{a+n}{b+n}\le\frac{a}{b}\) (nếu \(a\ge b\) )

\(\frac{a+n}{b+n}\ge\frac{a}{b}\) (nếu \(a\le b\) )