2/5 : x = 2 - 7/5

2/5 : x = 10/5 - 7/5

2/5 : x = 3/5

x = 2/5 : 3/5 

x = 2/3

Vậy x = 2/3

Bổ sung  x = 2

$\{\begin{array}{c}n+5⋮d\\ n+4⋮d\end{array}\iff d=1$

Vậy: n+5 và n+4 là hai số nguyên tố cùng nhau

\(\left\{101;110;200\right\}\)

\(\left[{}\begin{matrix}\dfrac{3}{2}-\dfrac{1}{2}x=0\\\dfrac{4}{3}x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{10}\end{matrix}\right.\)

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

\(\Leftrightarrow\dfrac{2x}{5}=\dfrac{14}{15}\)

\(\Leftrightarrow x=\dfrac{14}{15}\div\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{7}{3}\)

nhanh lên nhaaaaaaaaa

\(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{7.8.9.10}=\dfrac{3.119}{720.x}=\dfrac{119}{240.x}\)

\(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+\dfrac{6-3}{3.4.5.6}+...+\dfrac{10-7}{7.8.9.10}=\dfrac{119}{240.x}\)

\(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{3.4.5}-\dfrac{1}{4.5.6}+...+\dfrac{1}{7.8.9}-\dfrac{1}{8.9.10}=\dfrac{119}{240.x}\)

\(\dfrac{1}{1.2.3}-\dfrac{1}{8.9.10}=\dfrac{119}{240.x}\)

\(x=\left(\dfrac{1}{1.2.3}-\dfrac{1}{8.9.10}\right).\dfrac{240}{119}\)

`1+5+9+.....+195`

Khoảng cách : `4`

Số hạng:

(195-1)/4+1=49,5(số)`

`=>` Đề sai

______________________________

`5+8+11+....+29`

Khoảng cách :`3`

Số hạng:

`(29-5)/3 +1=9`(số)`

Tổng:

`(29+5)xx9:2=153`

Th1: 3/2 - 1/2x = 0

    =>  1/2x        = 3/2 - 0 = 3/2

   => x              = 3/2 : 1/2 = 3 

Th2: 4/3x + 2/5 = 0 

=> 4/3x              = 0-2/5 = -2/5

=>   x                    = -2/5 : 4/3 = -3/10 

A=1+3+3²+3³+...+3²⁰⁰⁰ ⋮ 13

  =(1+3+3²)+(3³+3⁴+3⁵)+...+(3¹⁹⁹⁸+3¹⁹⁹⁹+3²⁰⁰⁰) ⋮ 13

  =1.(1+3+3²)+3³.(1+3+3²)+...+3¹⁹⁹⁸.(1+3+3²) ⋮ 13

  =1.13+3³.13+...+3¹⁹⁹⁸.13 ⋮ 13

 =13.(1+3³+...+3¹⁹⁹⁸) ⋮ 13

Vì 13 ⋮ 13 nên 1+3+3²+3³+...+3²⁰⁰⁰ ⋮ 13

B=1+7+7²+7³+...+7²⁰¹⁸ ⋮ 19

  =(1+7+7²)+(7³+7⁴+7⁵)+...+(7²⁰¹⁶+7²⁰¹⁷+7²⁰¹⁸) ⋮ 19

  =1.(1+7+7²)+7³.(1+7+7²)+...+7²⁰¹⁶.(1+7+7²) ⋮ 19

  =1.57+7³.57+...+7²⁰¹⁶.57 ⋮ 19

  =57.(1+7³+...+7²⁰¹⁶) ⋮ 19

  Vì 57  ⋮ 19 nên 1+7+7²+7³+...+7²⁰¹⁸ ⋮ 19

\(A=1+3+3^2+...+3^{2000}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{1998}+3^{1999}+3^{2000}\right)\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1998}\left(1+3+3^2\right)\)

\(=13+13.3^3+...+13.3^{1998}\)

\(=13.\left(1+3^3+...+3^{1998}\right)⋮13\)

\(B=1+7+7^2+\left(7^3+7^4+7^5\right)+...+\left(7^{2016}+7^{2017}+7^{2018}\right)\)

\(=1+7+7^2+7^3\left(1+7+7^2\right)+...+7^{2016}\left(1+7+7^2\right)\)

\(=57+57.7^3+...+57.7^{2016}\)

\(=57.\left(1+7^3+...+7^{2016}\right)\)

Mà \(57⋮19\Rightarrow B⋮19\)

Câu C làm tương tự 2 câu trên (vẫn tách nhóm 3 số hạng)