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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\frac{n+3}{n+4}=\frac{(n+4)-1}{n+4}=1-\frac{1}{n+4}$
$\frac{n+1}{n+2}=\frac{(n+2)-1}{n+2}=1-\frac{1}{n+2}$
Vì $n+4> n+2$ nên $\frac{1}{n+4}< \frac{1}{n+2}$
Suy ra $1-\frac{1}{n+4}> 1-\frac{1}{n+2}$
Hay $\frac{n+3}{n+4}> \frac{n+1}{n+2}$
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$\frac{n-1}{n+4}< \frac{n-1}{n+2}=\frac{(n+2)-3}{n+2}=1-\frac{3}{n+2}$
$<1-\frac{n+3}=\frac{n}{n+3}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta so sánh hai phân số \(A=\frac{n}{n+3}\) và \(B=\frac{n-1}{n+4}\)
Ta thấy \(A+1=\frac{n}{n+3}+1=\frac{n}{n+3}+\frac{n+3}{n+3}=\frac{n+n+3}{n+3}=\frac{2n+3}{n+3}\)\(B+1=\frac{n-1}{n+4}+1=\frac{n-1}{n+4}+\frac{n+4}{n+4}=\frac{n-1+n+4}{n+4}=\frac{2n+3}{n+4}\)
Ta thấy \(2n+3=2n+3;n+3< n+4\Rightarrow\frac{2n+3}{n+3}>\frac{2n+3}{n+4}\Rightarrow A+1>B+1\Rightarrow A>B\)
Vậy \(\frac{n}{n+3}>\frac{n-1}{n+4}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{n+1}{n+2}=1-\frac{1}{n+2}\)
\(\frac{n+3}{n+4}=1-\frac{1}{n+4}\)
Mà \(\frac{1}{n+2}>\frac{1}{n+4}\)
Nên \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{n}{n+6}\)=\(1-\frac{6}{n+6}\)
\(\frac{n+1}{n+7}\)=\(1-\frac{6}{n+7}\)
Vì \(\frac{6}{n+6}>\frac{6}{n+7}\)=> \(\frac{n}{n+6}< \frac{n+1}{n+7}\)Vì phần cần thêm vào càng lớn thì phần có sẵn càng nhỏ
ủng hộ mik nhaaa
Ta có:
\(1-\frac{n}{n+6}=\frac{n+6}{n+6}-\frac{n}{n+6}=\frac{6}{n+6}.\)
\(1-\frac{n+1}{n+7}=\frac{n+7}{n+7}-\frac{n+1}{n+7}=\frac{6}{n+7}.\)
Vì \(n+6< n+7\)nên \(\frac{6}{n+6}>\frac{6}{n+7}\Leftrightarrow1-\frac{6}{n+6}< 1-\frac{6}{n+7}\Leftrightarrow\frac{n}{n+6}< \frac{n+1}{n+7}\)
k với!!!!!!!!!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{n+1}{n+2}=1-\frac{1}{n+2}\)
\(\frac{n+3}{n+4}=1-\frac{1}{n+4}\)
Mà \(\frac{1}{n+2}>\frac{1}{n+4}\)
Nne : \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)
\(\dfrac{n}{n+3}-\dfrac{n-1}{n+4}\)
\(=\dfrac{n^2+4n-n^2-2n+3}{\left(n+4\right)\left(n+3\right)}=\dfrac{2n+3}{\left(n+4\right)\left(n+3\right)}>0\)
=>n/n+3>(n-1)/(n+4)