Toán So sánh : $\frac{a}{b}$ và $\frac{a + 1}{b + 1}$ 11/07/2021 By Valentina So sánh : $\frac{a}{b}$ và $\frac{a + 1}{b + 1}$
`TH_1 :` `a<b`Ta có : `1-a/b=1/b` `1-(a+1)/(b+1)=1/(b+1)`Mà `1/b>1/(b+1)` nên `a/b<(a+1)/(b+1)``TH_2:` `a>b`Ta có : `a/b-1=1/b` `1-(a+1)/(b+1)=1/(b+1)`Mà `1/b>1/(b+1)` nên `a/b>(a+1)/(b+1)``TH_3:` `a=b``a/b=1=>(a+1)/(b+1)=1`Vậy `a/b=(a+1)/(b+1)` Trả lời
`TH1:a=b` `⇒a+1=b+1` $⇒\begin{cases}\dfrac{a}{b}=1\\\dfrac{a+1}{b+1}=1\end{cases}$ `⇒a/b={a+1}/{b+1}` `TH2:a>b` `⇒ab+a>ab+b` `⇒a(b+1)>b(a+1)` `⇒a/b>{a+1}/{b+1}` `TH3:a<b` `⇒ab+a<ab+b` `⇒a(b+1)<b(a+1)` `⇒a/b<{a+1}/{b+1}` Trả lời
`TH_1 :` `a<b`
Ta có : `1-a/b=1/b`
`1-(a+1)/(b+1)=1/(b+1)`
Mà `1/b>1/(b+1)` nên `a/b<(a+1)/(b+1)`
`TH_2:` `a>b`
Ta có : `a/b-1=1/b`
`1-(a+1)/(b+1)=1/(b+1)`
Mà `1/b>1/(b+1)` nên `a/b>(a+1)/(b+1)`
`TH_3:` `a=b`
`a/b=1=>(a+1)/(b+1)=1`
Vậy `a/b=(a+1)/(b+1)`
`TH1:a=b`
`⇒a+1=b+1`
$⇒\begin{cases}\dfrac{a}{b}=1\\\dfrac{a+1}{b+1}=1\end{cases}$
`⇒a/b={a+1}/{b+1}`
`TH2:a>b`
`⇒ab+a>ab+b`
`⇒a(b+1)>b(a+1)`
`⇒a/b>{a+1}/{b+1}`
`TH3:a<b`
`⇒ab+a<ab+b`
`⇒a(b+1)<b(a+1)`
`⇒a/b<{a+1}/{b+1}`