All Questions

jimin bts cute 08/08/2018 at 03:06
we put \(\left(a,b\right)=d\) inferred \(a=dm;b=d.n\) so in that \(\left(m,n\right)=1.\)
suppose \(a\le b\) then \(m\le n.\)
we have: \(ab=dm.dn=d^2m.n.\)
\(\left[a,b\right]=\dfrac{ab}{\left(a;b\right)}=\dfrac{d^2m.n}{d}=d.m.n\)
According to the post: \(\left[a,b\right]=210\) so \(d.m.n=210.\)
In that, \(d=\dfrac{ab}{\left[a,b\right]}=\dfrac{2940}{210}=14\) . So \(mn=\dfrac{210}{10}=15\)
We have following list:
\(m\) \(n\) \(a\) \(b\) \(1\) \(15\) \(14\) \(210\) \(3\) \(5\) \(42\) \(70\) 
jiminbts 08/08/2018 at 02:44
find 2 natural a and b with product of 2940 and smallest multiple is 210

jiminbts 08/08/2018 at 02:43
sorro mk dịch sang t.v nên nó không ra t.a
Quoc Tran Anh Le Coodinator
07/08/2018 at 05:36
Nguyễn Phương Uyên 07/08/2018 at 06:11
\(\dfrac{1}{4\dfrac{3}{7}}+\dfrac{1}{3\dfrac{11}{13}}+\dfrac{1}{\dfrac{5}{9}}\)
\(=\dfrac{1}{\dfrac{31}{7}}+\dfrac{1}{\dfrac{50}{13}}+\dfrac{1}{\dfrac{5}{9}}\)
\(=1\div\dfrac{31}{7}+1\div\dfrac{50}{13}+1\div\dfrac{5}{9}\)
\(=\dfrac{7}{31}+\dfrac{13}{50}+\dfrac{9}{5}\)
\(=\dfrac{3543}{1550}\)
Selected by MathYouLike 
Quoc Tran Anh Le Coodinator
07/08/2018 at 05:35Quoc Tran Anh Le Coodinator
07/08/2018 at 05:34
\(\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4}}}}=\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{\dfrac{17}{4}}}}=\dfrac{1}{4+\dfrac{1}{4+\dfrac{4}{17}}}=\dfrac{1}{4+\dfrac{1}{\dfrac{72}{17}}}=\dfrac{1}{4+\dfrac{17}{72}}=\dfrac{1}{\dfrac{305}{72}}=\dfrac{72}{305}\)
Selected by MathYouLike 
Nguyễn Phương Uyên 07/08/2018 at 06:14
\(\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4}}}}\)
\(=\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{\dfrac{17}{4}}}}\)
\(=\dfrac{1}{4+\dfrac{1}{4+1\div\dfrac{17}{4}}}\)
\(=\dfrac{1}{4+\dfrac{1}{4+\dfrac{4}{17}}}\)
\(=\dfrac{1}{4+\dfrac{1}{\dfrac{72}{17}}}\)
\(=\dfrac{1}{4+\dfrac{17}{72}}\)
\(=\dfrac{1}{\dfrac{140}{17}}\)
\(=\dfrac{17}{40}\)
Quoc Tran Anh Le Coodinator
07/08/2018 at 05:32
Lê Anh Duy 07/08/2018 at 09:55
Or if you mean 1 is the denominator, it will be different:
\(A=\dfrac{1}{\dfrac{\sqrt{.1}}{\sqrt{.01}}}=\dfrac{1^2}{\left(\dfrac{\sqrt{.1}}{\sqrt{.01}}\right)^2}=\dfrac{1}{\left(\dfrac{0.1}{0.01}\right)}=\dfrac{1}{10}=0.1\)
\(B=\dfrac{1}{\dfrac{\sqrt{.01}}{\sqrt{.1}}}=\dfrac{1^2}{\left(\dfrac{\sqrt{.01}}{\sqrt{.1}}\right)^2}=\dfrac{1}{\left(\dfrac{0.01}{0.1}\right)}=\dfrac{1}{0.1}=10\)
So A < B
BE CAREFUL WITH WHAT YOU WRITE!
Quoc Tran Anh Le selected this answer. 
Lê Anh Duy 07/08/2018 at 09:48
\(\dfrac{\dfrac{1}{\sqrt{.1}}}{\sqrt{.01}}=\dfrac{\left(\dfrac{1}{\sqrt{.1}}\right)^2}{\left(\sqrt{.01}\right)^2}=\dfrac{\dfrac{1}{0.1}}{0.01}=\dfrac{10}{0.01}=1000\)
\(\dfrac{\dfrac{1}{\sqrt{.01}}}{\sqrt{.1}}=\dfrac{\left(\dfrac{1}{\sqrt{.01}}\right)^2}{\left(\sqrt{.1}\right)^2}=\dfrac{\dfrac{1}{0.01}}{0.1}=\dfrac{100}{0.1}=1000\)
Answer: They are equal
Quoc Tran Anh Le Coodinator
07/08/2018 at 05:30
Lê Anh Duy 07/08/2018 at 10:34
First 3 fractions:
2 x 3 = 6 (the 5^{th})
3 x 3 = 9 (the 6^{th})
4 x 3 = 12 (the 7^{t}^{h})
Continue with the next 3 fractions:
5 x 6 = 30 (the 8^{th})
6 x 6 = 36 (the 9^{th})
9 x 6 = 54 (the 10^{th})
So \(\dfrac{1}{19}\) is incorrect and should be replaced with \(\dfrac{1}{30}\)
Quoc Tran Anh Le selected this answer.
Quoc Tran Anh Le Coodinator
07/08/2018 at 05:29Nguyễn Mạnh Hùng
07/08/2018 at 02:46Quoc Tran Anh Le Coodinator
07/08/2018 at 02:13
With this post, we have:
130%A = B
60%B = C
120%C = D
=> D = 60% of 120% of B
=> D = 72%B
Selected by MathYouLike