Đặt tông trên là A

\(\dfrac{2A}{7}=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)

\(\Rightarrow A=\dfrac{7.2022}{2.2023}=\dfrac{1011}{289}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2020}\)\(-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}\)

\(=\dfrac{2021}{2022}\)

Lời giải:

$\frac{1}{2^2}< \frac{1}{1.2}$

$\frac{1}{3^2}< \frac{1}{2.3}$

....

$\frac{1}{2016^2}< \frac{1}{2015.2016}$

Do đó:

$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}$

Đặt vế phải là $A$

$A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2015}-\frac{1}{2016}$

$=1-\frac{1}{2016}< 1$

Suy ra $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< A < 1$

\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016}\)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2016^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}\)<1 

Vậy ta có đpcm 

\(\dfrac{x+3}{x-1}=\dfrac{x-1+4}{x-1}=\dfrac{x-1}{x-1}+\dfrac{4}{x-1}=1+\dfrac{4}{x-1}\)

Để đạt GT nguyên thì \(\dfrac{4}{x-1}\in Z\)

\(\Rightarrow x-1\inƯ_{\left(4\right)}=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow x\in\left\{-3;-1;0;2;3;5\right\}\)

\(\dfrac{x-1+4}{x-1}=1+\dfrac{4}{x-1}\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

a

0,75x+4/9=-4/9

0,75x       = -4/9-4/9

0,75x       =  -4/9+(-4/9)

0,75x       =-8/9

       x       = -8/9:0,75

       x       =-8/9:75/100

       x       = -8/9:3/4

       x       =  -8/9.4/3

       x       =  -32/27   

Nam cho Bình `2/5` số bi thì Nam còn lại `:`

`1-2/5=3/5(` số bi `)`

Lúc đầu Nam có số bi là `:`

`12 : 3/5 = 20(` viên `)`

Đ/s `: 20` viên