\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{300}}\\ 3S=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{300}}\right)\\ 3S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{299}}\\ 3S-S=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{299}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{300}}\right)\\ 2S=1-\dfrac{1}{3^{300}}\\ S=\dfrac{1-\dfrac{1}{3^{300}}}{2}\)

Vậy \(S=\dfrac{1-\dfrac{1}{3^{300}}}{2}\)

\(\dfrac{1}{2}\) x \(\dfrac{4}{3}\) x 10 x \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\) = \(\dfrac{1}{2}\) x 2 = 1

(rút gọn 4/3 và 3/4 rồi rút 1/5 với 10, cuối cùng rút 1/2 và 2)

\(\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot...\cdot\dfrac{100}{99}\)

\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{10\cdot10}{9\cdot11}\)

\(=\dfrac{2\cdot3\cdot...\cdot10}{1\cdot2\cdot...\cdot9}\cdot\dfrac{2\cdot3\cdot...\cdot10}{3\cdot4\cdot...\cdot11}\)

\(=\dfrac{10}{1}\cdot\dfrac{2}{11}=\dfrac{20}{11}\)

\(\dfrac{2^{10}\cdot3^8-6^8}{4^4\cdot9^5}\)

\(=\dfrac{2^{10}\cdot3^8-2^8\cdot3^8}{2^8\cdot3^{10}}\)

\(=\dfrac{2^8\cdot3^8\left(2^2-1\right)}{2^8\cdot3^{10}}=\dfrac{1}{3^2}\cdot3=\dfrac{1}{3}\)

 M = \(\dfrac{18-4n}{n-3}\) (n \(\in\) Z)

M \(\in\) Z ⇔ 18 - 4n ⋮ n - 3

                6 - (4n - 12) ⋮ n - 3

                6 - 4.(n - 3) ⋮ n - 3

                6 ⋮ n - 3

           n - 3 \(\in\) Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}

           n  \(\in\) {-3; 0; 1; 2; 4; 5; 6; 9}

Vậy để M = \(\dfrac{18-4n}{n-3}\) có giá trị nguyên thì n \(\in\){-3; 0; 1; 2; 4; 5; 6; 9}

a: \(B=\left(-\dfrac{1}{2}\right)+\left(-\dfrac{1}{6}\right)+...+\left(-\dfrac{1}{90}\right)\)

\(=-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(1-\dfrac{1}{10}\right)=-\dfrac{9}{10}\)

b: \(D=\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{1}{2\cdot15}+\dfrac{13}{15\cdot4}\)

=>\(\dfrac{D}{7}=\dfrac{5}{2\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot28}\)

=>\(\dfrac{D}{7}=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{15}-\dfrac{1}{28}=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{13}{28}\)

=>\(D=\dfrac{13}{4}\)

Lời giải:

$A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}$

$> \frac{1}{1.2}+\frac{1}{3.4}=\frac{7}{12}$

Vậy ta có đpcm.

Lời giải:

4/7 quả bí nặng $1\frac{1}{2}$ kg = $\frac{3}{2}$ kg

Quả bi nặng: $\frac{3}{2}: \frac{4}{7}=2,625$ (kg)

Lời giải:

$S=3+3^2+3^3+3^4+....+3^{2024}$

$A=3+3^2+(3^3+3^4+3^5)+(3^6+3^7+3^8)+....+(3^{2022}+3^{2023}+3^{2024})$

$=12+3^3(1+3+3^2)+3^6(1+3+3^2)+.....+3^{2022}(1+3+3^2)$

$=12+(1+3+3^2)(3^3+3^6+....+3^{2022})$

$=12+13(3^3+3^6+....+3^{2022})$ chia 13 dư 12

Vậy $S$ không chia hết cho 13. Bạn xem lại đề.

\(S=3+3^2+3^3+3^4+...+3^{2024}\)

\(S=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2022}+3^{2023}+3^{2024}\right)\)

\(S=36+3^3.\left(3+3^2+3^3\right)+...+3^{2021}.\left(3+3^2+3^3\right)\)

\(S=36+3^3.36+...+3^{2021}.36\)

\(S=36.\left(1+3^3+...+3^{2021}\right)\)

Vì \(36⋮13\) nên \(36.\left(1+3^3+...+3^{2021}\right)⋮13\)

Vậy \(S⋮13\)

`#NqHahh`

11h40p thì kim giờ và kim phút sẽ nằm ở hai vị trí số 11 và số 8, cách nhau 3 số

=>Kim giờ và kim phút sẽ tạo ra góc 90 độ

90 độ nha