a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)

=>-10<x<-13/7

hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)

\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)

mà x là số nguyên

nên \(x\in\left\{-1;0\right\}\)

+ \(5N=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}\)

\(N=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{98}}+\dfrac{1}{5^{99}}\)

\(\Rightarrow4N=5N-N=1-\dfrac{1}{5^{99}}\)

\(\Rightarrow N=\dfrac{1}{4}-\dfrac{1}{4\cdot5^{99}}< \dfrac{1}{4}\) ( đpcm )

\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+........+\dfrac{1}{100^2}\)

Ta có :

\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)

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

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{6}{25}\)

Mà \(\dfrac{1}{6}< \dfrac{5}{26}< \dfrac{1}{4}\)

Mà \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+.........+\dfrac{1}{100^2}< \dfrac{6}{25}\)

\(\Leftrightarrow\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{100^2}< \dfrac{1}{4}\left(đpcm\right)\) \(\left(1\right)\)

a: 2x(x-1/7)=0

=>x(x-1/7)=0

=>x=0 hoặc x=1/7

b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)

c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)

\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)

\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)

mà x là số nguyên

nên \(x\in\left\{-4;-3;-2;-1\right\}\)

a: Gọi số nguyên cần tìm là x

Theo đề, ta có: \(\dfrac{1}{3}+\left(\dfrac{2}{4}-1\dfrac{2}{5}\right)< x< 2\dfrac{1}{7}+\left(\dfrac{-2}{5}-\dfrac{1}{4}\right)\)

\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{7}{5}< x< \dfrac{15}{7}-\dfrac{2}{5}-\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20}{60}+\dfrac{30}{60}-\dfrac{84}{60}< x< \dfrac{15\cdot20-2\cdot28-35}{140}\)

\(\Leftrightarrow-\dfrac{34}{60}< x< \dfrac{209}{140}\)

mà x là số nguyên

nên \(x\in\left\{0;1\right\}\)

b: Gọi số nguyên cần tìm là x

Theo đề, ta có: \(\dfrac{7}{3}+\dfrac{3}{4}-\dfrac{1}{5}>x>\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{7\cdot20+3\cdot15-12}{60}>x>\dfrac{56-21+2\cdot12}{84}\)

\(\Leftrightarrow\dfrac{173}{60}>x>\dfrac{59}{84}\)

mà x là số nguên

nên \(x\in\left\{2;1\right\}\)

b: \(\left|x-\dfrac{3}{5}\right|< \dfrac{1}{3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>-\dfrac{1}{3}\\x-\dfrac{3}{5}< \dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\dfrac{4}{15}< x< \dfrac{14}{15}\)

c: \(\left|x+\dfrac{11}{2}\right|>-5.5\)

mà \(\left|x+\dfrac{11}{2}\right|\ge0\forall x\)

nên \(x\in R\)

Ta có:

\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)

\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)

\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+\dfrac{4}{4!}-\dfrac{1}{4!}+...+\dfrac{100}{100!}-\dfrac{1}{100!}\)

\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+...+\dfrac{1}{99!}-\dfrac{1}{100!}\)

\(=1-\dfrac{1}{100!}\)

Mà \(1-\dfrac{1}{100!}< 1\)

Vậy \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) (Đpcm)

\(\dfrac{1}{2!}\)+ \(\dfrac{2}{3!}\)+ \(\dfrac{3}{4!}\)+...+\(\dfrac{99}{100!}\)

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

= 1-\(\dfrac{1}{100!}\) < 1.