A đâu !!

anh cũng đang định hỏi câu này

Ta có \(A=\frac{1}{2!}+\frac{2}{3!}+...+\frac{2014}{2015!}\)

=>  \(A=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2015-1}{2015!}\)

=>  \(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2015}{2015!}-\frac{1}{2015!}\)

=> \(A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2014!}-\frac{1}{2015!}\)

=>  \(A=1-\frac{1}{2015!}< 1\)

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

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

=>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.......

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 : \(\frac{n-1}{n+4}< 1\)=) \(\frac{n-1}{n+4}< \frac{n-1+1}{n+4+1}=\frac{n}{n+5}< \frac{n}{n+3}\)
=) \(\frac{n-1}{n+4}< \frac{n}{n+3}\)

h) Ta có: \(\frac{n+1}{n+2}=1-\frac{1}{n+2}\)

\(\frac{n+3}{n+4}=\frac{1}{n+4}\)

Vì \(n+2< n+4\)\(\Rightarrow\frac{1}{n+2}>\frac{1}{n+4}\)

\(\Rightarrow1-\frac{1}{n+2}< 1-\frac{1}{n+4}\)\(\Rightarrow\frac{n+1}{n+2}< \frac{n+3}{n+4}\)

a) Vì \(\frac{87}{39}>1\)

\(\frac{2015}{2017}< 1\)

\(\Rightarrow\frac{87}{39}>\frac{2015}{2017}\)

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

\(\Rightarrow\frac{n}{n+1}=\frac{n\cdot\left(n+3\right)}{\left(n+1\right)\left(n+3\right)}\)

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

\(\Rightarrow n\cdot\left(n+3\right)=n^2+3n\)

\(\Rightarrow\left(n+1\right)^2=n^2+2n+1\)

Dấu bằng chỉ xảy ra khi n = 1

Còn với mọi trường hợp n > 1 thì 

\(\frac{n}{n+1}>\frac{n+1}{n+3};n^2+3n>n^2+2n+1\)

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

\(\Rightarrow\frac{n}{n+3}=\frac{n\cdot\left(n+4\right)}{\left(n+3\right)\left(n+4\right)}\)

\(\Rightarrow n\cdot\left(n+4\right)=n^2+4n\)

\(\Rightarrow\left(n-1\right)\left(n+3\right)=n^2+2n-3\)

\(\Rightarrow n^2+4n>n^2+2n+3\)

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

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 : \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)