All Questions
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Nguyễn Mạnh Hùng 14/06/2018 at 02:22
We have:
\(8\sqrt{3}=\sqrt{8^2.3}=\sqrt{64.3}=\sqrt{192}\)
\(5\sqrt{7}=\sqrt{5^2.7}=\sqrt{25.7}=\sqrt{175}\)
Because 192>175 so \(\sqrt{192}>\sqrt{175}\)
or \(8\sqrt{3}>5\sqrt{7}\)
Lê Quốc Trần Anh selected this answer.
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\(n^5-n+2=n\left(n^4-1\right)+2=n\left(n^2+1\right)\left(n^2-1\right)+2\)
\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+2\)
We see: \(\left(n-1\right)\cdot n\cdot\left(n+1\right)⋮3\) (is product of 3 consecutive natural numbers)
\(\Rightarrow n^5-n+2\equiv2\left(mod3\right)\)
But no square numbers are in the form 3k + 2
So \(n\notin\varnothing\)
Lê Quốc Trần Anh selected this answer.
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quanpham04106 24/05/2018 at 02:24
because p,q,r is a prime number,pq+r=73 , ''the odd number'' multiply ''the odd number'' equal ''the odd number'' and ''the odd number'' plus ''the even number'' equal ''the odd number''. number.
We have 2 case:
case 1:r is even. so r = 2
pq+2=7
pq=71=1*71(eliminate)
case 2:p or q is even
example:p is even.so p=2
first:r=3
2*q+3=73
2*q=70
q=35
q isn't the prime number(eliminate)
second:r=5
2*q+5=73
2*q=68
q=34(eliminate)
third:r=7
2*q+7=73
2*q=66
q=33
fourth:r=9
2*q+9=73
2*q=64
q=32(eliminate)
fifth:r=11
2*q+11=73
2*q=62
q=31(possible)
so p+q+r=2+31+11=44
Lê Quốc Trần Anh selected this answer. -
Nguyễn Thành Long 26/05/2018 at 08:21
vì p, q, r là số nguyên tố, pq + r = 73, '' số lẻ '' nhân '' số lẻ '' bằng '' số lẻ '' và '' số lẻ '' cộng ' 'số chẵn' 'bằng' 'số lẻ' '. con số.
Chúng ta có 2 trường hợp:
trường hợp 1: r là chẵn. vậy r = 2
pq + 2 = 7
pq = 71 = 1 * 71 (loại bỏ)
trường hợp 2: p hoặc q là chẵn
ví dụ: p là even.so p = 2
đầu tiên: r = 3
2 * q + 3 = 73
2 * q = 70
q = 35
q không phải là số nguyên tố (loại bỏ)
thứ hai: r = 5
2 * q + 5 = 73
2 * q = 68
q = 34 (loại bỏ)
thứ ba: r = 7
2 * q + 7 = 73
2 * q = 66
q = 33
thứ tư: r = 9
2 * q + 9 = 73
2 * q = 64
q = 32 (loại bỏ)
thứ năm: r = 11
2 * q + 11 = 73
2 * q = 62
q = 31 (có thể)
do đó p + q + r = 2 + 31 + 11 = 44
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We have : \(2^{18}=262144=262121+23\)
Total time we can calculate since John was born until he was married is :
\(S=2^{18}\left(h\right)+1\left(h\right)=262144\left(h\right)+1\left(h\right)=262128\left(h\right)+17\left(h\right)\)
\(=10920\left(days\right)+17\left(h\right)=10922\left(days\right)+2\left(days\right)+17\left(h\right)\)
\(=1560\left(weeks\right)+2\left(days\right)+17\left(h\right)\)
Because after a week , the day that Fransisco was born was unchange so it still on Tuesday
After Tuesday 2 days is Thurday
So , exactly the day that Francisco was married is at 17:00 on Thursday
Lê Quốc Trần Anh selected this answer. -
huynh anh phuong 23/05/2018 at 10:28
218=131072 (hours)
Francisco get married at:131072+1=131073 hours
131073 :24 = 5461,375 days
The percent of one day is:24:100= 0,24%
Because 0,5 day = 12 hours so 6 hours=0,25 day.
And 0,375-0,25=0,125(day) so 0,125 day=7 hours.
Answer:5461 days 7 hours
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Apply inequality cauchy, we have:
\(\left\{{}\begin{matrix}\left(b+c-a\right)+\left(a+c-b\right)\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\\left(a+c-b\right)+\left(a+b-c\right)\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\\left(a+b-c\right)+\left(b+c-a\right)\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2c\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\2a\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\2b\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c^2\ge\left(b+c-a\right)\left(a+c-b\right)\\a^2\ge\left(a+c-b\right)\left(a+b-c\right)\\b^2\ge\left(a+b-c\right)\left(b+c-a\right)\end{matrix}\right.\)
\(\Rightarrow\left(abc\right)^2\ge\left[\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\right]^2\)
\(\Rightarrow abc\ge\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)
Lê Quốc Trần Anh selected this answer. -
Fc Alan Walker 18/05/2018 at 13:41
Apply inequality cauchy, we have:
⎧⎩⎨⎪⎪⎪⎪(b+c−a)+(a+c−b)≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√(a+c−b)+(a+b−c)≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√(a+b−c)+(b+c−a)≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√
⇒⎧⎩⎨⎪⎪⎪⎪2c≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√2a≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√2b≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√
⇒⎧⎩⎨⎪⎪c2≥(b+c−a)(a+c−b)a2≥(a+c−b)(a+b−c)b2≥(a+b−c)(b+c−a)
⇒(abc)2≥[(b+c−a)(a+c−b)(a+b−c)]2
⇒abc≥(b+c−a)(a+c−b)(a+b−c)
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Choose the smallest number in this list is a, and b = a+1, c = a+2, d = a+3
So abcd + 1 = a(a+1)(a+2)(a+3) + 1 = a(a+3).(n+1)(n+2) + 1
= (a2 + 3a)(n2 + 3n +2) + 1
= [a2 + 3a][(n2 + 3n) + 2] + 1
= (a2 + 3a)2 + 2.(a2 + 3a).1 + 12 = (a2 + 3a + 1)2 is a square number
Lê Quốc Trần Anh selected this answer.
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I think you are wrong in the number of even and odd numbers in English
even numbers: số chẵn odd numbers: số lẻ
So in this post must be n is a odd number.
\(n^4-10n^2+9=n^4-n^2-9n^2+9=n^2\left(n^2-1\right)-9\left(n^2-1\right)=\left(n^2-1\right)\left(n^2-9\right)=\left(n-1\right)\left(n+1\right)\left(n-3\right)\left(n+3\right)\)
Because n is a odd number so we can write n at the form n = 2k+1
So \(\left(n-1\right)\left(n+1\right)\left(n-3\right)\left(n+3\right)=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1-3\right)\left(2k+1+3\right)\)
\(=2k\left(2k+2\right)\left(2k-2\right)\left(2k+4\right)=2k\cdot2\left(k+1\right)\cdot2\left(k-1\right)\cdot2\left(k+2\right)=16\left(k-1\right)k\left(k+1\right)\left(k+2\right)⋮16\)
But (k-1)k(k+1)(k+2) is the product of four consecutive integers so has multiples of 2,3,4 so \(⋮\left(2\cdot3\cdot4\right)=24\)
So \(n^4-10n^2+9⋮\left(16.24\right)=384\)
Lê Quốc Trần Anh selected this answer.
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Huy Toàn 8A (TL) 21/05/2018 at 02:06
B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)
We have:
4B = 1.2.3.4 + 2.3.4.4 + ... + (n - 1)n(n + 1).4
= 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n - 1)n(n + 1)(n + 2) - [(n - 2)(n - 1)n(n + 1)]
= (n - 1)n(n + 1)(n + 2) - 0.1.2.3 = (n - 1)n(n + 1)(n + 2)
=>B = \(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)
Nguyễn Thị Linh selected this answer. -
Nguyễn Thành Long 26/05/2018 at 08:34
Ta có :
\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)
\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right).n.\left(n+1\right).4\)
\(4B=1.2.3.4+2.3.4\left(5-1\right)+...+\left(n-1\right).n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)
\(4B=1.2.3.4+2.3.4.5-1.2.3.4+...+n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\)
\(4B=\left[1.2.3.4+2.3.4.5+...+n.\left(n+1\right).\left(n+2\right)\right]\)\(-\left[1.2.3.4+2.3.4.5+...+\left(n-1\right).n.\left(n+1\right)\right]\)
\(4B=n.\left(n+1\right).\left(n+2\right)\)
\(B=\dfrac{n.\left(n+1\right).\left(n+2\right)}{4}\)