\(\dfrac{1}{x}-\dfrac{y}{11}=-\dfrac{2}{11}\)

\(\dfrac{1}{x}=-\dfrac{2}{11}+\dfrac{y}{11}\)

\(\dfrac{1}{x}=\dfrac{y-2}{11}\)

\(x\left(y-2\right)=11\)

\(\Rightarrow x,\left(y-2\right)\inƯ\left(11\right)=\left\{1,-1,11,-11\right\}\)

có bảng sau :

Vậy ...

\(\dfrac{1}{x}-\dfrac{y}{11}=-\dfrac{2}{11}\Rightarrow11-xy=-2x\)

\(\Leftrightarrow-2x+xy=11\Leftrightarrow x\left(-2+y\right)=11\)

\(\Rightarrow x;y-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

A=111​+121​+...+701​

\(A = \left(\right. \frac{1}{11} + \frac{1}{12} + . . . + \frac{1}{20} \left.\right) + \left(\right. \frac{1}{21} + \frac{1}{22} + . . . + \frac{1}{30} \left.\right)\)

\(+ \left(\right. \frac{1}{31} + \frac{1}{32} + . . . + \frac{1}{40} \left.\right) + \left(\right. \frac{1}{41} + \frac{1}{42} + . . . + \frac{1}{50} \left.\right) + \left(\right. \frac{1}{51} + \frac{1}{52} + . . . + \frac{1}{60} \left.\right)\)

\(+ \left(\right. \frac{1}{61} + \frac{1}{62} + . . . + \frac{1}{70} \left.\right)\)

\(\Rightarrow A < \frac{1}{10} \cdot 10 + \frac{1}{20} \cdot 10 + \frac{1}{30} \cdot 10 + . . . + \frac{1}{60} \cdot 10\)

\(A < 1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{6}\)

\(A < 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \left(\right. \frac{1}{4} + \frac{1}{5} \left.\right)\)

\(A < 2 + 0 , 45 < 2 , 5\)

A= 11 1 ​ + 12 1 ​ +...+ 70 1 ​ A = ( 1 11 + 1 12 + . . . + 1 20 ) + ( 1 21 + 1 22 + . . . + 1 30 ) A=( 11 1 ​ + 12 1 ​ +...+ 20 1 ​ )+( 21 1 ​ + 22 1 ​ +...+ 30 1 ​ ) + ( 1 31 + 1 32 + . . . + 1 40 ) + ( 1 41 + 1 42 + . . . + 1 50 ) + ( 1 51 + 1 52 + . . . + 1 60 ) +( 31 1 ​ + 32 1 ​ +...+ 40 1 ​ )+( 41 1 ​ + 42 1 ​ +...+ 50 1 ​ )+( 51 1 ​ + 52 1 ​ +...+ 60 1 ​ ) + ( 1 61 + 1 62 + . . . + 1 70 ) +( 61 1 ​ + 62 1 ​ +...+ 70 1 ​ ) ⇒ A < 1 10 ⋅ 10 + 1 20 ⋅ 10 + 1 30 ⋅ 10 + . . . + 1 60 ⋅ 10 ⇒A< 10 1 ​ ⋅10+ 20 1 ​ ⋅10+ 30 1 ​ ⋅10+...+ 60 1 ​ ⋅10 A < 1 + 1 2 + 1 3 + . . . + 1 6 A<1+ 2 1 ​ + 3 1 ​ +...+ 6 1 ​ A < 1 + 1 2 + 1 3 + 1 6 + ( 1 4 + 1 5 ) A<1+ 2 1 ​ + 3 1 ​ + 6 1 ​ +( 4 1 ​ + 5 1 ​ ) A < 2 + 0 , 45 < 2 , 5 A<2+0,45<2,5

Đây qu, phiền bạn tick giup mình nha

Ta thấy : \(\frac{5}{11}>\frac{5}{12}>\frac{5}{13}>\frac{5}{14}\)

 \(S=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}< \frac{5}{11}\times4=\frac{20}{11}< 2\)  (1)

\(S=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}>\frac{5}{14}\times4=\frac{10}{7}>1\)   (2)

Từ (1) và (2) suy ra : \(1< S< 2\)  (ĐPCM)

S=​​​\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}<\frac{4}{10}+\frac{4}{10}+\frac{4}{10}+\frac{4}{10}+\frac{4}{10}\)

                                                                     =\(\frac{4}{10}\cdot5=2=>S<2\)

       S=\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}<\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}\)

=\(\frac{3}{15}\cdot5=1=>S>1\)

Vậy 1<S<2

nhớ k với nhé

Đặt  \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2140.2141}\)

Có \(\frac{1}{2^3}< \frac{1}{2.3};\frac{1}{3^3}< \frac{1}{3.4};...;\frac{1}{2140^3}< \frac{1}{2140.2141}\)

\(\Rightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2140^3}< A\). Từ đó ta tính được A

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2140}-\frac{1}{2141}\)

\(A=\frac{1}{2}-\frac{1}{2141}\Rightarrow A>\frac{1}{2}\). Mà \(\frac{1}{2}< \frac{2}{3}\Rightarrow A< \frac{2}{3}\)

Có \(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2140^3}< A\Rightarrow\)\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2140^3}< \frac{2}{3}\)

a: 7/8>7/10

b: 16/5>16/7

c: 8/7>1

d: 15/11>1

e: 4/9<1<9/4

f: 11/10>1>10/11

x\(⋮\)13 =) x \(\in\)B ( 13 ) = { 0 ; +-13 ; +-26 ; +-39 }

nhưng -14<x<27 

=) x \(\in\){ 0 ; +-13 ; 26 }