Question

so sánh

a\(\frac{n}{n+1}\)và \(\frac{n+2}{n+3}\)

b \(\frac{n}{n+3}\)và \(\frac{n-1}{n+4}\)

c \(\frac{n}{2n+1}\)và\(\frac{3n+1}{6n+3}\)

Answer 1

a). n/n+1  < n+2/n+3 

b). n/n+3 > n−1/n+4 

c). n/2n+1 < 3n+1/6n+3 

k mk nha

Answer 2

\(\frac{n}{n+1}< 1\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+1+2}=\frac{n+2}{n+3}\)

=>n/n+1<n+2/n+3

vậy........

b)\(\frac{n}{n+3}>\frac{n}{n+4}>\frac{n-1}{n+4}\Rightarrow\frac{n}{n+3}>\frac{n}{n+4}\)

vậy.....

c)\(\frac{n}{2n+1}=\frac{3n}{6n+3}< \frac{3n+1}{6n+3}\)

vậy.......

Answer 3

a) \(\frac{n}{n+1}=1-\frac{1}{n+1};\frac{n+2}{n+3}=1-\frac{1}{n+3}\)
Vì \(\frac{1}{n+1}>\frac{1}{n+3}\)=) \(1-\frac{1}{n+1}< 1-\frac{1}{n+3}\)
=) \(\frac{n}{n+1}< \frac{n+2}{n+3}\)
b) Áp dụng tính chất : Nếu \(\frac{a}{b}< 1\)=) \(\frac{a}{b}< \frac{a+m}{b+m}\)
 Ta có : \(\frac{n-1}{n+4}< 1\)=) \(\frac{n-1}{n+4}< \frac{n-1+1}{n+4+1}=\frac{n}{n+5}< \frac{n}{n+3}\)
=) \(\frac{n-1}{n+4}< \frac{n}{n+3}\)

Answer 4

a) \(\frac{n}{n+1}v\text{à}\frac{n+2}{n+3}\)

ta có : \(\frac{n}{n+1}=\frac{\left(n+1\right)-1}{n+1}=1-\frac{1}{n+1}\)

\(\frac{n+2}{n+3}=\frac{\left(n+3\right)-1}{n+3}=1-\frac{1}{n+3}\)

ta có \(\frac{1}{n+1}>\frac{1}{n+3}\Rightarrow1-\frac{1}{n-1}< 1-\frac{1}{n+3}\)

\(\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+3}\)