\(\dfrac{x+1}{3-x}< 0\left(x\ne3\right)\\ =>-\dfrac{x+1}{x-3}< 0\\ =>\dfrac{x+1}{x-3}>0\\ =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\\ =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x< 3\end{matrix}\right.\end{matrix}\right.\\ =>\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)

`1/[5.6]-1/[6.7]-1/[7.8]-...-1/[3.2]-1/[2.1]`

`=1/5-1/6-1/6+1/7-1/7+1/8-...-1/3+1/2-1/2+1`

`=1/5-1/3+1`

`=13/15`

Sửa: `4/[3.7]+4/[7.11]+...+4/[95.99]`

`=1/3-1/7+1/7-1/11+....+1/95-1/99`

`=1/3-1/99`

`=32/99`

A = \(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + ......+ \(\dfrac{4}{95.99}\)

A = \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + ......+ \(\dfrac{1}{95}\) - \(\dfrac{1}{99}\)

A = \(\dfrac{1}{3}\) - \(\dfrac{1}{99}\)

A = \(\dfrac{32}{99}\)

Lời giải:
$A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2022^2}$

$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2021.2022}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2021}-\frac{1}{2022}$

$=1-\frac{1}{2022}< 1$

Vậy $A< 1$

0,25 = \(\dfrac{1}{4}\) 

vậy 0,25 là một số hữu tỉ 

có