tìm số nguyên dương n nhỏ nhất để các phân số này tối giản 3/n+7;4/n+8;5/n+9;...;15/n+19
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\dfrac{2}{3}:x=2,4-\dfrac{4}{5}\)
=>\(\dfrac{2}{3}:x=2,4-0,8=1,6\)
=>\(x=\dfrac{2}{3}:1,6=\dfrac{2}{4,8}=\dfrac{1}{2,4}=\dfrac{5}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{-5}{6}\cdot\dfrac{14}{19}+\dfrac{-9}{12}\cdot\dfrac{14}{19}-\dfrac{5}{18}\)
\(=\dfrac{14}{19}\left(-\dfrac{5}{6}-\dfrac{9}{12}\right)-\dfrac{5}{18}\)
\(=\dfrac{14}{19}\cdot\dfrac{-10-9}{12}-\dfrac{5}{18}\)
\(=\dfrac{14}{19}\cdot\dfrac{-19}{12}-\dfrac{5}{18}=\dfrac{-7}{6}-\dfrac{5}{18}\)
\(=\dfrac{-26}{18}=-\dfrac{13}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=3+\dfrac{3}{5}+\dfrac{3}{5^2}+...+\dfrac{3}{5^9}\)
=>\(5S=15+3+\dfrac{3}{5}+...+\dfrac{3}{5^8}\)
=>\(5S-S=15+3+...+\dfrac{3}{5^8}-3-\dfrac{3}{5}-...-\dfrac{3}{5^9}\)
=>\(4S=15-\dfrac{3}{5^9}=\dfrac{15\cdot5^9-3}{5^9}\)
=>\(S=\dfrac{15\cdot5^9-3}{4\cdot5^9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=1+\dfrac{1}{2}+...+\dfrac{1}{2^{100}}\)
=>\(2C=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)
=>\(2C-C=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}-1-\dfrac{1}{2}-...-\dfrac{1}{2^{100}}\)
=>\(C=2-\dfrac{1}{2^{100}}=\dfrac{2^{101}-1}{2^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=\dfrac{1}{2^x}\)
=>\(\dfrac{2}{2}\cdot\dfrac{3}{6}\cdot\dfrac{4}{8}\cdot...\cdot\dfrac{30}{60}\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=\dfrac{1}{2^x}\)
=>\(\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{2^x}\)
=>\(\dfrac{1}{2^{29}}\cdot\dfrac{1}{2^6}=\dfrac{1}{2^x}\)
=>x=29+6=35
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(A=\dfrac{3}{10}+\dfrac{3}{40}+...+\dfrac{3}{340}\)
\(=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 10:
Số học sinh giỏi ngoại ngữ chiếm:
\(\dfrac{1}{3}:\dfrac{4}{5}=\dfrac{1}{3}\cdot\dfrac{5}{4}=\dfrac{5}{12}\)(tổng số học sinh)
Số học sinh giỏi Văn là:
\(1-\dfrac{1}{3}-\dfrac{5}{12}=\dfrac{12-4-5}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)(tổng số học sinh)
Tổng số học sinh là: \(6:\dfrac{1}{4}=24\left(bạn\right)\)
Số học sinh giỏi toán là \(24\cdot\dfrac{1}{3}=8\left(bạn\right)\)
Số học sinh giỏi ngoại ngữ là 24-8-6=10(bạn)
Bài 11:
a: Để A là phân số thì \(x+2\ne0\)
=>\(x\ne-2\)
b: Để A là số nguyên thì \(2x-1⋮x+2\)
=>\(2x+4-5⋮x+2\)
=>\(-5⋮x+2\)
=>\(x+2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-1;-3;3;-7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{2}{5}\cdot4-\dfrac{3}{5}\cdot7\cdot\dfrac{3}{2}\)
\(=\dfrac{8}{5}-\dfrac{63}{10}\)
\(=\dfrac{16}{10}-\dfrac{63}{10}=-\dfrac{47}{10}\)
b: \(-\dfrac{21}{10}+\dfrac{21}{10}\cdot\dfrac{3}{4}-\dfrac{3}{4}\)
\(=\dfrac{-21}{10}\left(1-\dfrac{3}{4}\right)-\dfrac{3}{4}\)
\(=-\dfrac{21}{10}\cdot\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-21}{40}-\dfrac{3}{4}=\dfrac{-51}{40}\)
c: \(\dfrac{-2}{5}\cdot\left(-6\right)+\dfrac{3}{4}\cdot\dfrac{4}{-10}\)
\(=\dfrac{12}{5}-\dfrac{3}{10}\)
\(=\dfrac{24}{10}-\dfrac{3}{10}=\dfrac{21}{10}\)
a)2/5.4-3/5.7.3/2 b)-21/10 + 21/10 . 3/4-34
=8/5-63/10 =-21/10 +(21/10 . 3/4) -3/4
=-47/10 =-21/10+63/40-3/4
=-56/40