\(B=\frac{\frac{1}{39}-\frac{1}{6}-\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}:\frac{31}{6}\)

\(=\frac{\frac{1}{3}\left(\frac{1}{13}-\frac{1}{2}-\frac{1}{17}\right)}{\frac{1}{4}\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}.\frac{6}{31}\)

\(=\frac{\frac{-1}{3}\left(\frac{-1}{13}+\frac{1}{2}+\frac{1}{17}\right)}{\frac{1}{4}\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}.\frac{6}{31}\)

\(=\frac{-1}{3}:\frac{1}{4}.\frac{6}{31}\)

\(=\frac{-1}{3}.4.\frac{6}{31}\)

Tiếp theo dễ r tự làm tiếp :)

a, Vì  A, B < 1

\(A=\frac{15^{16}+1}{15^{17}+1}< \frac{15^{16}+1+14}{15^{17}+1+14}=\frac{15^{16}+15}{15^{17}+15}=\frac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}\)

b, \(B=\frac{2018^{2018}+1}{2018^{2019}+1}< 1< \frac{2018^{2019}+1}{2018^{2018}+1}=A\)

\(d=\left(21a+4,14a+3\right)\Rightarrow\hept{\begin{cases}21a+4⋮d\\14a+3⋮d\end{cases}}\Rightarrow\hept{\begin{cases}42a+8⋮d\\42a+9⋮d\end{cases}}\Rightarrow\left(42a+9\right)-\left(42a+8\right)=1⋮d\Rightarrow d=1\) 

\(\Rightarrow\text{đ}cpm\)

Gọi \(\left(21n+4;14n+3\right)=d\)

\(\Rightarrow\hept{\begin{cases}21n+4⋮d\\14n+3⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2.\left(21n+4\right)⋮d\\3.\left(14n+3\right)⋮d\end{cases}\Rightarrow\hept{\begin{cases}42n+8⋮d\\42n+9⋮d\end{cases}}}\)

\(\Rightarrow\left(42n+9\right)-\left(42n+8\right)⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Rightarrow\frac{21n+4}{14n+3}\)là phân số tối giản với mọi n là số tự nhiên

\(|x+1|+|x+2|+|x+3|=4x-4\left(1\right)\)

Vì \(\hept{\begin{cases}|x+1|\ge0;\forall x\\|x+2|\ge0;\forall x\\|x+3|\ge0;\forall x\end{cases}}\)

\(\Rightarrow|x+1|+|x+2|+|x+3|\ge0;\forall x\)

Mà \(|x+1|+|x+2|+|x+3|=4x-4\)

\(\Rightarrow4x-4\ge0\)

\(\Rightarrow4x\ge4\)

\(\Rightarrow x\ge1>0\)

\(\Rightarrow\hept{\begin{cases}|x+1|=x+1\\|x+2|=x+2\\|x+3|=x+3\end{cases}}\)(2) 

Thay (2) vào (1) ta được :

\(x+1+x+2+x+3=4x-4\)

\(\Leftrightarrow3x+6=4x-4\)

\(\Leftrightarrow3x-4x=-4-6\)

\(\Leftrightarrow-x=-10\)

\(\Leftrightarrow x=10\)

Vậy x=10

\(C=3+3^2+3^3+...+3^{100}\)

\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+3^{97}\left(1+3+3^2+3^3\right)\)

\(=\left(1+3+3^2+3^3\right)\left(3+3^5+...+3^{97}\right)\)

\(=40\left(3+3^5+...+3^{97}\right)⋮40\left(đpcm\right)\)

C = 3 + 3² + 3⁴ + ... + 3¹⁰⁰

   = (3 + 3²) + (3⁴ + 3⁶) + ... + (3⁹⁸ + 3¹⁰⁰)

   = 3(1 + 3) + 3⁴(1 + 3²) + ... + 3⁹⁸(1 + 3²)

   = 3.4 + 3⁴.10 + ... + 3⁹⁸.10

   = 3.4 + 10(3⁴ + ... + 3⁹⁸)

Ta có: \(\hept{\begin{cases}3.4⋮4\\10\left(3^4+...+3^{98}\right)⋮10\end{cases}}\)=> C \(⋮\)40 (đpcm)

a) A = {x\(\in\)N | x = ab ; b - a = 4} ( a\(\ne\)0)

b) B = { x \(\in\)N | x = ab ; a + b = 4 } ( a \(\ne\)0)

c) C = {x\(\in\)N | x = ab; a,b \(\in\){0;1;2} (a \(\ne\)0)

d) D = { x\(\in\)N| x = ab; a ,b \(\in\){3;4;5} ; ab \(⋮̸\)2}