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We have:

\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)=2\) (*)

Other way:

\(a^2+ab+b^2=a^2+2.\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

\(=\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

Because \(\left(a+\dfrac{1}{2}b\right)^2\ge0\) with \(\forall a,b\)

\(\dfrac{3}{4}b^2\ge0\) with \(\forall b\)

So \(\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\ge0\) with \(\forall a,b\)

or \(a^2+ab+b^2\ge0\) with \(\forall a,b\) (**)

From (*) and (**) we have  \(a+b\le2\)

Your ex is so hard to do :) 

With x=2 then \(4P\left(2\right)=5.2^2=20\Rightarrow P\left(2\right)=5\)

With x=3 then \(P\left(3\right)+3P\left(2\right)=5.3^2=45\Rightarrow P\left(3\right)=45-5.3=30\)

\(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{zx}{80}=\dfrac{xy+yz+zx}{24+30+80}=\dfrac{1206}{134}=9\)

\(\Rightarrow x^2=64.9=8^2.3^2\Rightarrow x=\pm24\)

With x=24 then y=24:8.3=9;z=30

With x=-24 then y=-9;z=-30

We have:

\(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\)

\(\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}\)

Apply the same sequence properties, we have:

\(\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}=\dfrac{xy+yz+xz}{24+30+80}=\dfrac{1206}{134}=9\)

So \(x^2=9.64=576\)\(\Rightarrow x=24\)

\(\dfrac{xy}{24}=9\Rightarrow xy=24.9=216\Rightarrow y=9\)

\(\dfrac{yz}{30}=9\Rightarrow yz=270\Rightarrow z=30\)

So \(\left(x;y;z\right)=\left(24;9;30\right)\)

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 \(8\sqrt{3}>5\sqrt{7}\) 

\(n^5-n+2=n\left(n^4-1\right)+2=n\left(n^2+1\right)\left(n^2-1\right)+2\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+2\)

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 \(n\notin\varnothing\)

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

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

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 

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

Apply inequality cauchy, we have:

\(\left\{{}\begin{matrix}\left(b+c-a\right)+\left(a+c-b\right)\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\\left(a+c-b\right)+\left(a+b-c\right)\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\\left(a+b-c\right)+\left(b+c-a\right)\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2c\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\2a\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\2b\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}c^2\ge\left(b+c-a\right)\left(a+c-b\right)\\a^2\ge\left(a+c-b\right)\left(a+b-c\right)\\b^2\ge\left(a+b-c\right)\left(b+c-a\right)\end{matrix}\right.\)

\(\Rightarrow\left(abc\right)^2\ge\left[\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\right]^2\)

\(\Rightarrow abc\ge\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)

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)

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

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 \(n^4-10n^2+9⋮\left(16.24\right)=384\)

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 = \(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

Ta có :           

\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right).n.\left(n+1\right).4\)

\(4B=1.2.3.4+2.3.4\left(5-1\right)+...+\left(n-1\right).n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)

\(4B=1.2.3.4+2.3.4.5-1.2.3.4+...+n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\)

\(4B=\left[1.2.3.4+2.3.4.5+...+n.\left(n+1\right).\left(n+2\right)\right]\)\(-\left[1.2.3.4+2.3.4.5+...+\left(n-1\right).n.\left(n+1\right)\right]\)

\(4B=n.\left(n+1\right).\left(n+2\right)\)

\(B=\dfrac{n.\left(n+1\right).\left(n+2\right)}{4}\)

We have:\(x:y:z=a:b:c\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\)

\(\Rightarrow\left(x+y+z\right)^2=\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\)

We have 

\(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{b}{c}\Leftrightarrow\dfrac{10a+b}{10b+c}=\dfrac{b}{c}=\dfrac{10a+b-b}{10b+c-c}=\dfrac{10a}{10b}=\dfrac{a}{b}\)

=> \(\dfrac{a}{b}=\dfrac{b}{c}\)

64 squares