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1 ) 10 \(⋮\) n
=> n \(\in\) Ư ( 10 )
Ư ( 10 ) = { 1 , 2 , 5 , 10 }
Vậy n \(\in\) { 1 ; 2 ; 5 ; 10 }
2 ) 12 : \(⋮\) ( n - 1 )
=> n - 1 \(\in\) Ư ( 12 )
=> Ư ( 12 ) = { 1 ; 12 ; 2 ; 6 ; 3 ; 4 }
n - 1 | 1 | 12 | 2 | 6 | 3 | 4 |
n | 2 | 13 | 3 | 7 | 4 | 5 |
Vậy n \(\in\) { 2 , 13 , 3 , 7 , 4 , 5 }
3 ) 20 \(⋮\) ( 2n + 1 )
=> 2n + 1 \(\in\) Ư ( 20 )
=> Ư ( 20 ) = { 1 ; 20 ; 2 ; 10 ; 4 ; 5 }
2n+1 | 1 | 20 | 2 | 10 | 4 | 5 |
n | 0 | 19/2 ( loại ) | 1/2 ( loại ) | 9/2 ( loại ) | 3/2 ( loại ) | 2 |
Các trường hợp loại , vì n \(\in\) N
Vậy n thuộc { 0 , 2 }
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1+1/A+1/a2+1/a3+1+.../an+1
=1(1/A/a2/a3/...an)
=1.(1/a1+2+3+...+n)
=1.(1/a6+...+n)
=a6+...+n
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\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
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a)\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{100.101}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{100}\)-\(\frac{1}{101}\)=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)
b)\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{28.29.30}\)=\(\frac{868}{3480}\)=\(\frac{217}{870}\)
c)\(\frac{1}{1.2.3.4}\)+\(\frac{1}{2.3.4.5}\)+....+\(\frac{1}{27.28.29.30}\)=\(\frac{24354}{438480}\)=\(\frac{451}{8120}\)
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\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\\ 7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+....+\frac{1}{69}-\frac{1}{70}\right)\\ 7.\left(\frac{1}{10}-\frac{1}{70}\right)\\ 7.\frac{6}{70}=\frac{3}{5} \)
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\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)=-\frac{3}{20}\)
Đặt \(A=\) \(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(\Rightarrow A=\left(-1\right)\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
Đặt \(S=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(S=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(S=\frac{1}{4}-\frac{1}{10}\)
\(S=\frac{3}{20}\)
\(\Rightarrow A=\left(-1\right)\frac{3}{20}=\frac{-3}{20}\)
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\(\frac{429}{639}>\frac{219}{333}\)
\(\frac{10^{15}+1}{10^{16}+1}>\frac{10^{16}+1}{10^{17}+1}\)
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a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\)
\(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\)
\(\frac{3}{7}x=\frac{22}{35}\)
\(x=\frac{49}{35}=1,4\)
b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\)
\(x:\frac{1}{2}=1,5-\frac{1}{4}\)
\(x:\frac{1}{2}=\frac{5}{4}\)
\(x=\frac{5}{4}.\frac{1}{2}\)
\(x=\frac{5}{8}\)
Vậy ..
1+1=2 : )
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