\(\left\{{}\begin{matrix}x⋮12\\x⋮21\end{matrix}\right.\) =>\(x\in BC\left(12;21\right)\)

Ta có: 12= \(2^2\cdot3\)

            21 = 3*7

=> BCNN(12;21)= 2² *3*7= 84

BC(12;21)=B(84)={0;84;168;252;336;420;504;....}

Vì 18<x<500 =>x\(\in\left\{84;168;252;336;420\right\}\)

\(B=\dfrac{4}{8\cdot13}+\dfrac{4}{13\cdot18}+\dfrac{4}{18\cdot23}+...+\dfrac{1}{253\cdot258}\)

\(B=4\left(\dfrac{1}{8\cdot13}+\dfrac{1}{13\cdot18}+\dfrac{1}{18\cdot23}+...+\dfrac{1}{253\cdot258}\right)\)

\(B=\dfrac{4}{5}\left(\dfrac{5}{8\cdot13}+\dfrac{5}{13\cdot18}+\dfrac{5}{18\cdot23}+...+\dfrac{5}{253\cdot258}\right)\)

\(B=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)

\(B=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)\)

\(B=\dfrac{4}{5}\cdot\dfrac{125}{1032}=\dfrac{25}{258}\)

\(\left(2x+1\right)^4=\left(2x+1\right)^2\)

\(\Rightarrow\left(2x+1\right)^4-\left(2x+1\right)^2=0\)

\(\left(2x+1\right)^2\left[\left(2x+1\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=0\\\left(2x+1\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\)

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

\(S=2+2^2+2^3+...+2^{10}\) 

\(S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\) (5 cặp )

\(S=2\left(1+2\right)+2^3\left(1+2\right)+..+2^9\left(1+2\right)\)

\(S=2\cdot3+2^3\cdot3+...+2^9\cdot3\)

\(=3\left(2+2^3+..+2^9\right)⋮3\left(dpcm\right)\)

\(3+3^2+3^3+3^4\)

\(=3\left(1+3+3^2+3^3\right)\)

\(=3\cdot40\)

\(=3\cdot10\cdot4⋮4\left(dpcm\right)\)

\(2+2^2+2^3+....+2^{2022}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2021}+2^{2022}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2021}\left(1+2\right)\)

\(=2\cdot3+2^3\cdot3+...+2^{2021}\cdot3\)

\(=3\left(2+2^3+...+2^{2011}\right)⋮3\left(dpcm\right)\)

Vì số chia là 3 nên dư có thể là 1 hoặc 2 

Lại có dư là 1 số lẻ nên số dư là 1 

Vậy số cần tìm:

3*15+1 =46

\(A=0+2+4+...+1000\) (501 số hạng)

\(2A=\left(0+1000\right)+\left(2+998\right)+\left(4+996\right)+...+\left(1000+0\right)\) (501 CẶP

\(2A=1000\cdot501\)

\(A=\dfrac{1000\cdot501}{2}=250500\)

Gọi số cần tìm là \(\overline{abc}\)

Theo bài ra ta có:

\(\overline{abc6}-\overline{abc}=6063\)

\(\overline{abc0}+6-\overline{abc}=6063\)

\(10\cdot\overline{abc}+6-\overline{abc}=6063\)

\(9\cdot\overline{abc}+6=6063\)

\(9\cdot\overline{abc}=6057\)

=> \(\overline{abc}=6057:9=673\)

Vậy số cần tìm là 673

Số học sinh giỏi là :

\(42\cdot\dfrac{2}{3}=28\) (học sinh)

Số học sinh trung bình là :

\(41\cdot\dfrac{1}{6}=7\) (học sinh)

Vậy số học sin yếu là :

42-28-7=7(học sinh)