Question

So sánh bt N là số tự nhiên:

\(\dfrac{n+3}{n+4}\)và,\(\dfrac{n+1}{n+2}\)                      \(\dfrac{n-1}{n+4}\) và  \(\dfrac{n}{n+3}\)

Answer 1

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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$\frac{n-1}{n+4}< \frac{n-1}{n+2}=\frac{(n+2)-3}{n+2}=1-\frac{3}{n+2}$

$<1-\frac{n+3}=\frac{n}{n+3}$