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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!

\(\dfrac{1}{2}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=\dfrac{1}{2^x}\)

=>\(\dfrac{2}{2}\cdot\dfrac{3}{6}\cdot\dfrac{4}{8}\cdot...\cdot\dfrac{30}{60}\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=\dfrac{1}{2^x}\)

=>\(\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{2^x}\)

=>\(\dfrac{1}{2^{29}}\cdot\dfrac{1}{2^6}=\dfrac{1}{2^x}\)

=>x=29+6=35

?

a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

\(=1-\dfrac{1}{6}=\dfrac{5}{6}\)

b: \(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{10100}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

c: \(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

d: \(A=\dfrac{3}{10}+\dfrac{3}{40}+...+\dfrac{3}{340}\)

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

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

\(=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)