\(x = {6 {} \over 11}\)

\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{10000}\right)\)

\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)

\(=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)

\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

k nha ruiminh giai 

Ta có:  \(\frac{1}{4}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{9}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{16}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

.............................................................

\(\frac{1}{10000}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)

\(\Rightarrow\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

\(\Rightarrow\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}< 1\)(đpcm)

Xin tk

Ta có: $\frac{1}{4}$+$\frac{1}{9}$+$\frac{1}{16}$+.....+$\frac{1}{10000}$=$\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$

mà $\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$ < $\frac{1}{1.2}$+$\frac{1}{2.3}$+.......+$\frac{1}{99.100}$

$\Rightarrow$ $\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$ < 1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+.......+$\frac{1}{99}$-$\frac{1}{100}$

$\Rightarrow$ $\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$ < 1-$\frac{1}{100}$

$\Rightarrow$ $\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$ < 1

xl nhe em mình nó cứ vào copy phá nick của mình

S=1/4+1/9+1/16+...+1/10000 = 1/2x 2 + 1/3x3+...+1/100x100 < 1/1x2 + 1/2x3 +...+ 1/9x10 = 1 - 1/2 + 1/2 - 1/3 +...+ 1/9 - 1/10 = 1- 1/10 < 1

S=1/4+1/9+1/16+...+1/10000

 = 1/2x 2 + 1/3x3+...+1/100x100 < 1/1x2 + 1/2x3 +...+ 1/9x10

= 1 - 1/2 + 1/2 - 1/3 +...+ 1/9 - 1/10 = 1- 1/10 < 1

A=1/(2x2)+1/(3x3)+...+1/(100x100) 
Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) 
=> A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4

A=1/(2x2)+1/(3x3)+...+1/(100x100) Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) => A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4