Tham khảo
`ĐK:a,b \ne 0`
Giải sử `\frac{a+n}{b+n}=\frac{a}{b}`
`⇒(a+n)b=(b+n)a`
`⇒ab+bn=ab+an`
`⇒bn=an`
`⇒b=a`
Giả sử `\frac{a+n}{b+n}>\frac{a}{b}`
`⇒(a+n)b>(b+n)a`
`⇒ab+bn>ab+an`
`⇒bn>an`
`⇒b>a`
Giả sử `\frac{a+n}{b+n}<\frac{a}{b}`
`⇒(a+n)b<(b+n)a`
`⇒ab+bn<ab+an`
`⇒bn<an`
`⇒b<a`
Vậy `b=a⇒\frac{a+n}{b+n}=\frac{a}{b}`
`b>a⇒\frac{a+n}{b+n}>\frac{a}{b}`
`b<a⇒\frac{a+n}{b+n}<\frac{a}{b}`
`\text{©CBT}`