Question

Cho a,b,c \(\inℕ^∗\). So sánh \(\frac{a+n}{b+n}\)và \(\frac{a}{b}\)

Answer 1

Mình lộn chữ "c" sửa thành chữ "n" nha

Answer 2

ta có:a<b
1-a+n/b+n =(b+n-a-n)/a+n=>(b-a)/a+n
Vì (b-a)/a < (b-a)/a+n nên a/b ( b>0) > a+n/b+n
Làm tương tự Vs a>b nha!

Answer 3

#)Giải :

Ta xét ba trường hợp :

- Trường hợp 1 : Nếu a > b

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

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

Vì \(\frac{a-b}{b}>\frac{a-b}{b+n}\Rightarrow\frac{a}{b}-1>\frac{a+n}{b+n}-1\)hay \(\frac{a}{b}>\frac{a+n}{b+n}\)khi a > b

- Trường hợp 2 : Nếu a < b

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

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

Vì \(\frac{b-a}{b}>\frac{b-a}{b+n}\Rightarrow1-\frac{a}{b}>1-\frac{a+n}{b+n}\)hay \(\frac{a}{b}< \frac{a+n}{b+n}\)khi a < b

- Trường hợp 3 : Nếu a = b

\(\Rightarrow\frac{a+n}{b+n}=\frac{a}{b}=1\) 

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

Answer 4

\(TH1:a< b\)

\(\Rightarrow an< bn\)

\(\Rightarrow an+ab< bn+ab\)

\(\Rightarrow a\left(b+n\right)< b\left(a+n\right)\)

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

\(TH2:a=b\)

\(\Rightarrow an=bn\Rightarrow an+ab=bn+bn\)

\(\Rightarrow a\left(b+n\right)=b\left(a+n\right)\Rightarrow\frac{a}{b}=\frac{a+n}{b+n}\)

\(TH3:a>b\)

\(\Rightarrow an>bn\Rightarrow an+ab>bn+ab\)

\(\Rightarrow a\left(b+n\right)>b\left(a+n\right)\)

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