Mình mới lớp 5 nên không biết làm bài này.

Xin lỗi nha! Chúc bạn may mắn......mình chính là Đào Minh Tiến!

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

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

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

So sánh : \(n\cdot\left(n+3\right)\)và \(\left(n+2\right)\cdot\left(n+3\right)\)

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

\(\left(n+2\right)\cdot\left(n+3\right)=n^2+5n+6\)

\(n^2+3n< n^2+5n+6\)

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

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

\(\frac{n}{2n+1}=\frac{n\cdot\left(6n+3\right)}{\left(2n+1\right)\cdot\left(6n+3\right)}\)

\(\frac{3n+1}{6n+3}=\frac{\left(3n+1\right)\cdot\left(2n+1\right)}{\left(6n+3\right)\cdot\left(2n+1\right)}\)

So sánh : \(n\cdot\left(6n+3\right)\)và \(\left(3n+1\right)\cdot\left(2n+1\right)\)

\(n\cdot\left(6n+3\right)=6n^2+3n\)

\(\left(3n+1\right)\cdot\left(2n+1\right)=6n^2+5n+1\)

\(6n^2+3n< 6n^2+5n+1\)

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

n/n+3=n:(n+3)=n:n+n:3=1+n:3

n+1/n+2=(n+1):(n+2)=(n+1):n+(n+1):(n+2)=1+n+n/2+1/2=3/2+3n/2=3(1+n):2

Vì ta thấy rõ 3(1+n):2 > 1+n :3 

Hay n/n+3 < n+1/n+2

Ta xét 2 phân số sau thì có :

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

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

Để so sánh 2 phân số trên ta so sánh\(\frac{3}{n+3};\frac{1}{n+2}\)

Quy đồng lên ta có :

\(\frac{3}{n+3}=\frac{3\left(n+2\right)}{\left(n+3\right)\left(n+2\right)}=\frac{3n+6}{\left(n+3\right)\left(n+2\right)}\)

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

Mà 3n+6>n+3

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

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

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

Mọi người làm vại chi cho phức tạp.

Ta có: \(\frac{n}{n+3}< \frac{n}{n+2}\) (vì \(n+3>n+2\))

Và \(\frac{n}{n+2}< \frac{n+1}{n+2}\)(vì \(n< n+1\))

Theo tính chất bắc cầu suy ra \(\frac{n}{n+3}< \frac{n+1}{n+2}\)

Ta số phân số chung gian là \(\frac{n+1}{n+3}\)

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

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

Ủng hộ nhé ! 

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

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

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

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

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

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

\(\frac{n}{n+1}\)<\(\frac{n+2}{n+3}\) với n>=0 

Ta có : 

\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3}\)

....

\(\frac{1}{n^2}=\frac{1}{n.n}<\frac{1}{\left(n-1\right).n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}<1\)nên \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<1\)