So sánh : \(\frac{a+n}{b+n}\) và \(\frac{a}{b}\)
Vì \(\frac{a+n}{b+n}=\frac{a}{b}\) \(\Rightarrow\frac{a}{b}=\frac{a}{b}\)
Vậy \(\frac{a+n}{b+n}=\frac{a}{b}\)
Nếu a < b thì : \(\dfrac{a}{b}< 1\)
Giả sử \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\)
\(\Leftrightarrow a\left(b+n\right)< b\left(a+n\right)\)
\(\Leftrightarrow ab+an< ab+bn\)
\(\Leftrightarrow an< bn\)
\(\Leftrightarrow a< b\)
\(\Rightarrow\dfrac{a}{b}< 1\) ( luôn đúng )
Nếu a = b thì \(\dfrac{a}{b}=\dfrac{a+n}{b+n}=1\)
Nếu a > b thì \(\dfrac{a}{b}>1\)
Giả sử \(\dfrac{a}{b}>\dfrac{a+n}{b+n}\)
\(\Leftrightarrow a\left(b+n\right)>b\left(a+n\right)\)
\(\Leftrightarrow ab+an>ab+bn\)
\(\Leftrightarrow an>bn\)
\(\Leftrightarrow a>b\)
\(\Rightarrow\dfrac{a}{b}>1\) ( luôn đúng )
=> ( đpcm )
Ta xét các trường hợp:
* \(\frac{a}{b}\)<1
\(\Rightarrow\) a < b
\(\Rightarrow\)an < bn ( Nhân 2 vế cho n > 0 )
\(\Rightarrow\)an+ab < bn+ab ( cộng hai vế cho ab )
\(\Rightarrow\)a. ( n+b ) < b. ( n+a )
\(\Rightarrow\)\(\frac{a}{b}\) < \(\frac{a+n}{b+n}\)
* \(\frac{a}{b}\)> 1
\(\Rightarrow\)a > b
\(\Rightarrow\)an > bn
\(\Rightarrow\)an+ab > bn+ab
\(\Rightarrow\)a. ( n+b ) > b. ( n+a )
\(\Rightarrow\)\(\frac{a}{b}\)> \(\frac{a+n}{b+n}\)
Vậy ta có hai đáp án là
\(\frac{a}{b}\)<\(\frac{a+n}{b+n}\) và \(\frac{a}{b}\) > \(\frac{a+n}{b+n}\)
Nếu:
\(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a+n}{b+n}>1\Leftrightarrow\dfrac{a+n}{b+n}< \dfrac{a}{b}\)
Nếu
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a+n}{b+n}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\)
Xét 3 trường hợp :
+ TH1: a = b (a, b \(\in\) N*)
=> \(\left\{{}\begin{matrix}a+n=b+n=>\dfrac{a+n}{b+n}=1\\\dfrac{a}{b}=1\end{matrix}\right.\)
=> \(\dfrac{a}{b}=\dfrac{a+n}{b+n}\)
+ TH2: a > b (a, b \(\in\) N*)
Ta có : \(\dfrac{a}{b}=\dfrac{a\left(b+n\right)}{b\left(b+n\right)}=\dfrac{ab+an}{b^2+bn}\) (n \(\in\) N*)
\(\dfrac{a+n}{b+n}=\dfrac{\left(a+n\right)b}{\left(b+n\right)b}=\dfrac{ab+bn}{b^2+bn}\)
Ta có : a > b => an > bn ( vì n \(\in\) N*)
=> ab + an > ab + bn
=> \(\dfrac{ab+an}{b^2+bn}>\dfrac{ab+bn}{b^2+bn}\)
=> \(\dfrac{a}{b}>\dfrac{a+n}{b+n}\)
+ TH3: a < b (a, b \(\in\) N*)
Ta có : \(\dfrac{a}{b}=\dfrac{a\left(b+n\right)}{b\left(b+n\right)}=\dfrac{ab+an}{b^2+bn}\) (n \(\in\) N*)
\(\dfrac{a+n}{b+n}=\dfrac{\left(a+n\right)b}{\left(b+n\right)b}=\dfrac{ab+bn}{b^2+bn}\)
Ta có : a < b => an < bn ( vì n \(\in\) N*)=> ab + an < ab + bn
=> \(\dfrac{ab+an}{b^2+bn}< \dfrac{ab+bn}{b^2+bn}\)
=> \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\)
Vậy \(\dfrac{a}{b}=\dfrac{a+n}{b+n}\)khi a = b \(\dfrac{a}{b}>\dfrac{a+n}{b+n}\)khi a > b \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\)khi a < b
