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help e câu d) vs ạ '_'

\(P=\dfrac{1}{2022}+\dfrac{2}{2021}+\dfrac{3}{2020}+...+\dfrac{12}{2011}+\dfrac{13}{1}\\ P=\left(\dfrac{1}{2022}+1\right)+\left(\dfrac{2}{2021}+1\right)+.....+\left(\dfrac{12}{2011}+1\right)+1\\ P=\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{2011}+\dfrac{2023}{2023}\\ P=2023\cdot\left(\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{2022}+\dfrac{1}{2023}\right)\)

\(\dfrac{P}{S}=\dfrac{2023\cdot\left(\dfrac{1}{2023}+\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{2011}\right)}{\dfrac{1}{2011}+\dfrac{1}{2012}+...+\dfrac{1}{2023}}\\ =2023\)

Help e câu d bài 7 vs ạ

\(2x^5+2=4\\ 2x^5=2\\ x^5=1\\ x=1\)

a, x ∈ B(12) 

B(12) = {0;12;24;36;48;60;84;96;108;.........}

Vì \(x\le100\)

=> \(x\in\left\{\text{0;12;24;36;48;60;84;96}\right\}\)

b, x ∈ Ư(20)

Ư(20) = {1;2;5;4;10;20}

Vì x > 8 

=> x ∈ {10;20}

c. x ∈ B(8)

B(8) = {0;8;16;24;32;40;48;56;64;72;80;.......;120;128}

Vì \(12\le x\le120=>x\in\left\{16;24;32;......;120\right\}\)

d. x ∈ Ư(45)

Ư(45) = {1;5;3;9;15;45}

vì x > 5 => x ∈ {9;15;45}

\(b.75\%-1\dfrac{2}{3}+0,5:\dfrac{5}{12}\)

\(=\dfrac{3}{4}-\dfrac{5}{3}+\dfrac{6}{5}\)

\(=\dfrac{17}{60}\)

\(d.\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+....+\dfrac{1}{399}+\dfrac{1}{483}\)

\(=\left(\dfrac{2}{3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+.....+\dfrac{2}{21\cdot19}+\dfrac{2}{21\cdot23}\right)\cdot\dfrac{1}{2}\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\right)\cdot\dfrac{1}{2}\)

\(=\left(1-\dfrac{1}{23}\right)\cdot\dfrac{1}{2}\)

\(=\dfrac{22}{23}\cdot\dfrac{1}{2}\)

\(=\dfrac{11}{23}\)

3. 

\(C=5+5^2+...+5^6\\ C=5\left(1+5\right)+5^3\left(1+5\right)+5^5\left(1+5\right)\\ C=5\cdot6+5^3\cdot6+5^5\cdot6\\ C=6\left(5+5^3+5^5\right)\\ =>C⋮6\)

\(4.D=3+3^2+...+3^6\\D=3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)\\ D=3\cdot4+3^3\cdot4+3^5\cdot5\\ C=4\left(3+3^3+3^5\right)\\ =>D⋮4\\ 5.E=4+4^2+...+4^6\\ E=4\left(1+4\right)+4^3\left(1+4\right)+4^5\left(1+4\right)\\ E=4\cdot5+4^3\cdot5+4^5\cdot5\\ E=5\left(4+4^3+4^5\right)=>E⋮5 \)