9: \(A=\dfrac{3^2}{10}+\dfrac{3^2}{40}+...+\dfrac{3^2}{340}\)

\(=3\left(\dfrac{3}{10}+\dfrac{3}{40}+...+\dfrac{3}{340}\right)\)

\(=3\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\right)\)

\(=3\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(=3\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=3\cdot\dfrac{9}{20}=\dfrac{27}{20}\)

10: \(A=\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)

\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{25}-\dfrac{1}{31}\right)\)

\(=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)

11: \(A=\dfrac{6}{15}+\dfrac{6}{35}+\dfrac{6}{63}+\dfrac{6}{99}\)

\(=3\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\)

\(=3\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}\right)\)

\(=3\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)

\(=3\left(\dfrac{1}{3}-\dfrac{1}{11}\right)=3\cdot\dfrac{8}{33}=\dfrac{8}{11}\)

12: \(A=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{49\cdot51}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)=\dfrac{3}{2}\cdot\dfrac{16}{51}=\dfrac{8}{17}\)

13: \(A=\dfrac{1}{2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\)

\(=1-\dfrac{1}{16}=\dfrac{15}{16}\)

14: \(A=\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}\)

15: \(A=\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{294}+\dfrac{8}{609}\)

\(=\dfrac{3}{6\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{7}{14\cdot21}+\dfrac{8}{21\cdot29}\)

\(=\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{29}\)

\(=\dfrac{1}{6}-\dfrac{1}{29}=\dfrac{23}{174}\)

16: \(A=\dfrac{5}{24}+\dfrac{5}{104}+\dfrac{5}{234}+\dfrac{5}{414}\)

\(=\dfrac{5}{3\cdot8}+\dfrac{5}{8\cdot13}+\dfrac{5}{13\cdot18}+\dfrac{5}{18\cdot23}\)

\(=\dfrac{1}{3}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}\)

\(=\dfrac{1}{3}-\dfrac{1}{23}=\dfrac{20}{69}\)

17: \(A=\dfrac{\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{294}}{\dfrac{5}{24}+\dfrac{5}{104}+\dfrac{5}{234}}\)

\(=\dfrac{\dfrac{1}{6}-\dfrac{1}{21}}{\dfrac{1}{3}-\dfrac{1}{18}}=\dfrac{15}{126}:\dfrac{15}{54}=\dfrac{54}{126}=\dfrac{3}{7}\)

\(A=\dfrac{5\cdot4^6\cdot9^4-3^9\cdot\left(-8\right)^4}{4\cdot2^{13}\cdot3^8+2\cdot8^4\cdot\left(-27\right)^3}\)

\(=\dfrac{5\cdot2^{12}\cdot3^8-3^9\cdot2^{12}}{2^{15}\cdot3^8-2^{13}\cdot3^9}\)

\(=\dfrac{2^{12}\cdot3^8\left(5-3\right)}{2^{13}\cdot3^8\left(2^2-3\right)}\)

\(=\dfrac{2^{13}}{2^{13}}=1\)

MỌI NGƯỜI GIÚP MK VS Ạ

 rảnh à

??????????????????????????

số học sinh của trường có thể là 280 hoặc 501.

Gọi số học sinh của trường là x(bạn)

(Điều kiện: \(x\in Z^+\))

Số học khi xếp hàng 13 thì dư 4 em nên \(x-4\in B\left(13\right)\)

=>\(x-4\in\left\{...;247;260;273;...;598;...\right\}\)

=>\(x\in\left\{...;251;264;277;...;602;...\right\}\)

mà 250<=x<=600

nên \(x\in\left\{251;264;277;...;589\right\}\left(1\right)\)

Số học sinh khi xếp hàng 17 thì dư 9 em nên \(x-9\in B\left(17\right)\)

=>\(x-9\in\left\{...;255;272;...;595;...\right\}\)

=>\(x\in\left\{...;264;281;...;604;...\right\}\)

mà 250<=x<=600

nên \(x\in\left\{264;281;...;587\right\}\left(2\right)\)

Số học sinh khi xếp hàng 5 thì vừa hết nên \(x\in B\left(5\right)\)

mà 250<=x<=600

nên \(x\in\left\{250;255;260;...;600\right\}\left(3\right)\)

Từ (1),(2),(3) suy ra 

\(\left\{{}\begin{matrix}x\in\left\{251;264;...;589\right\}\\x\in\left\{264;281;...;587\right\}\\x\in\left\{250;255;260;...;600\right\}\end{matrix}\right.\)

=>x=485

Vậy: Số học sinh là 485 bạn

1: \(\left(-12,5\right)+17,55+\left(-3,5\right)-\left(-2,45\right)\)

\(=\left(-12,5-3,5\right)+17,55+2,45\)

=-16+20

=4

2: \(\dfrac{-3}{5}\cdot\dfrac{2}{7}+2\dfrac{3}{5}-\dfrac{3}{5}\cdot\dfrac{5}{7}\)

\(=-\dfrac{3}{5}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{13}{5}\)

\(=-\dfrac{3}{5}+\dfrac{13}{5}=\dfrac{10}{5}=2\)

3: \(\dfrac{2}{3}:x=2,4-\dfrac{4}{5}\)

=>\(\dfrac{2}{3}:x=2,4-0,8=1,6\)

=>\(x=\dfrac{2}{3}:1,6=\dfrac{2}{4,8}=\dfrac{1}{2,4}=\dfrac{5}{12}\)

\(\dfrac{-5}{6}\cdot\dfrac{14}{19}+\dfrac{-9}{12}\cdot\dfrac{14}{19}-\dfrac{5}{18}\)

\(=\dfrac{14}{19}\left(-\dfrac{5}{6}-\dfrac{9}{12}\right)-\dfrac{5}{18}\)

\(=\dfrac{14}{19}\cdot\dfrac{-10-9}{12}-\dfrac{5}{18}\)

\(=\dfrac{14}{19}\cdot\dfrac{-19}{12}-\dfrac{5}{18}=\dfrac{-7}{6}-\dfrac{5}{18}\)

\(=\dfrac{-26}{18}=-\dfrac{13}{9}\)

\(S=3+\dfrac{3}{5}+\dfrac{3}{5^2}+...+\dfrac{3}{5^9}\)

=>\(5S=15+3+\dfrac{3}{5}+...+\dfrac{3}{5^8}\)

=>\(5S-S=15+3+...+\dfrac{3}{5^8}-3-\dfrac{3}{5}-...-\dfrac{3}{5^9}\)

=>\(4S=15-\dfrac{3}{5^9}=\dfrac{15\cdot5^9-3}{5^9}\)

=>\(S=\dfrac{15\cdot5^9-3}{4\cdot5^9}\)

Cái này tính nhanh nhé!

\(C=1+\dfrac{1}{2}+...+\dfrac{1}{2^{100}}\)

=>\(2C=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)

=>\(2C-C=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}-1-\dfrac{1}{2}-...-\dfrac{1}{2^{100}}\)

=>\(C=2-\dfrac{1}{2^{100}}=\dfrac{2^{101}-1}{2^{100}}\)

Cái này tính nhanh nhé!