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1. 12 = \(6+5+\dfrac{3}{3}\)
2. 12 = \(3+3+6\)
3. 12 = \(\sum\limits^6_3\left(X\right)-6\)
4. 12 = \(\prod\limits^6_5\left(X\right)-3\cdot5-3\)
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\(4^{50}=2^{100}=2^{10}.2^{10}=\left(...4\right).\left(...4\right)=...6\)
So .......
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How about using Mincopsky's Inequality ?
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Condiction : x \(\ne\) 1
Following FA FIFA's answer , at the sixth row , we have :
\(\dfrac{3x-3}{x-1}=0\)
\(\Leftrightarrow\dfrac{3\left(x-1\right)}{x-1}=0\)
\(\Leftrightarrow3\left(x-1\right)=0\)
<=> x = 1
Impossible because for equation exists , x must different to 1
It also mean that there is not any roots satisfy equation .
P/s : FIFA's answer was wrong because a few equation you cannot compact the fraction . In equation , you must to use this sign "<=>" to indicate an equivalent equation !
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+) With a = b = c = 0 (satisfy)
+) Consider a,b,c \(\ne\) 0
-) With a,b,c are three odd numbers
=> (a2 + b2 + c2) / 8 (balance 3)
(Because with x is an odd number , square of x divided for 8 will balance 1, we can prove it if we use formula of odd numbers)
By the way , at right hand side (a2.b2) / 8 (balance 1)
It's contradictory
So with a,b,c are three odd numbers (removed)
-) With a,b,c are three even numbers
=> We decided \(\left\{{}\begin{matrix}a^2=2^n.x\\b^2=2^m.y\\c^2=2^p.z\end{matrix}\right.\)
=> \(a^2b^2\equiv0\) (mod 2m + n) <=> \(a^2b^2⋮2^{m+n}\)
But the left hand side is not
So with a,b,c are three even numbers (removed)
-) With a, b, c is not the same kind , we have this table
a b c odd even odd even odd odd Change the kind of three , it's not satisfy
So a = b = c = 0
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We have :
\(S=\left(x+y+1\right)\left(x^2+y^2\right)+\dfrac{4}{x+y}=\left(x+y\right)\left(x^2+y^2\right)+\left(x^2+y^2\right)+\dfrac{4}{x+y}\)
Applying Cauchy's inequality for two positive number , we have
\(S\ge2.\sqrt{\left(x+y\right)\left(x^2+y^2\right).\dfrac{4}{x+y}}+x^2+y^2=2.\sqrt{4\left(x^2+y^2\right)}+\left(x^2+y^2\right)\)
\(=4.\sqrt{\left(x^2+y^2\right)}+\left(x^2+y^2\right)=\sqrt{x^2+y^2}.\left(\sqrt{x^2+y^2}+5\right)\)
\(\ge1.\left(1+4\right)=5\)
So MinS = 5 <=> x = y = 1
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Because n is a positive number so n has two types
+) With n is an odd number (Such as: a1 ; a3 ; .....)
=> \(a_n=\left(-1\right).\dfrac{n^2+n+1}{n!}\)
Because a1 , a2 , a3 , ...... an are continuous numbers
So (n - 1) is an even number
=> \(a_{n-1}=1.\dfrac{n^2+n+1}{n!}\)
With 2 continuous numbers, we can see , the total of 2 number equals to 0
\(\left(-1\right).\dfrac{n^2+n+1}{n!}+1.\dfrac{n^2+n+1}{n!}=0\)
So , with 2017 continuous number , S's value is
S = \(\left(a_1+a_2\right)+\left(a_2+a_3\right)+......+\left(a_{2015}+a_{2016}\right)+a_{2017}\)
S = \(a_{2017}=\left(-1\right)^{2017}.\dfrac{2017^2+2017+1}{2017!}=-\dfrac{2017^2+2018}{2017!}\)
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\(\dfrac{x+b}{x-5}+\dfrac{x+5}{x-b}=2\)
\(\Rightarrow\dfrac{x+b-\left(x-5\right)}{x-5}+\dfrac{x+5-\left(x-b\right)}{x-b}=0\)
\(\Rightarrow\dfrac{b+5}{x-5}+\dfrac{b+5}{x-b}=0\)
\(\Rightarrow\left(b+5\right)\left(\dfrac{1}{x-5}+\dfrac{1}{x-b}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b+5=0\\\dfrac{1}{x-5}+\dfrac{1}{x-b}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=-5\\x-b+x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=-5\\x=\dfrac{b+5}{2}\end{matrix}\right.\)
+) With b = -5
=> \(\dfrac{x-5}{x-5}+\dfrac{x+5}{x+5}=2\) (Its true)
So with b = -5 and x \(\in R\) then equation is satisfy
+) With \(x=\dfrac{b+5}{2}\)
=> \(\dfrac{\dfrac{b+5}{2}+b}{\dfrac{b+5}{2}-5}+\dfrac{\dfrac{b+5}{2}+5}{\dfrac{b+5}{2}-b}=\dfrac{\dfrac{3b+5}{2}}{\dfrac{b-5}{2}}+\dfrac{\dfrac{b+15}{2}}{\dfrac{-b+5}{2}}=\dfrac{3b+5}{b-5}+\dfrac{b+15}{5-b}\)
\(=\dfrac{3b+5-\left(b+15\right)}{b-5}=\dfrac{2b-10}{b-5}=2\) (It's true)
So the pairs of numbers satisfying the equation (x;b) are \(\left(\infty;-5\right)\) ; \(\left(\dfrac{b+5}{2};\infty\right)\)
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a) Apply properties of certain segments , we have :
AD // BM
AB // MD
=> AD = BM
Similar , we have EA = MC
=> AD + EA = BM + MC
=> DE = BC (1)
We can see : \(\angle EDM=\angle ABC\) (Because ED // BM , AD // MB)
\(\angle DEM=\angle ACB\) (Because ED // BM , EM // AC)
=> \(\Delta MED=\Delta ACB\left(e-a-e\right)\)
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With x = 0
=> A = 0 + 0 + 6 = 6 (Removed) Because \(6⋮\left(1;2;3;6\right)\)
With x = 1
=> A = 1 + 1 + 6 = 8 (Removed) Because \(8⋮\left(1;2;4;8\right)\)
Similar , with x = 2 then A still not a prime number
With x \(\ge\) 3 => x has three forms
\(\left\{{}\begin{matrix}x=3k\\x=3k+1\\x=3k+2\end{matrix}\right.\)
+) With x = 3k => A = 9k2 + 3k + 6 = 3.(3k2 + k + 2) (Removed)
+) With x = 3k + 1 => A = (3k + 1)2 + (3k + 1) + 6 = 9k2 + 6k + 1 + 3k + 1 + 6 = 9k2 + 9k + 8
A = 9k(k + 1) + 8
We can see : k(k + 1) \(⋮\) 2 because it is a multiplication of 2 continuous numbers
8 \(⋮\) 2 too
So A \(⋮\) 2 => x = 3k + 1 (removed)
+) With x = 3k + 2 => A = (3k + 2)2 + (3k + 2) + 6 = 9k2 + 12k + 4 + 3k + 2 + 6 = 9k2 + 15k +12
A = 3.(3k2 + 5k + 4) \(⋮\) 3 (Removed)
So , without any integer value of x that satisfy A is a prime number
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Denote \(\left\{{}\begin{matrix}A=\dfrac{2017^x-2016^{y+1}}{2015}\\B=2017^x-2016^{y+1}\\C=2015\end{matrix}\right.\)
For A is a square number then \(B⋮C\)
<=> \(2017^x-2016^{y+1}⋮2015\)
<=> \(2017^x-2^x+2^x-2016^{y+1}+1-1⋮2015\)
<=> \(2^x-1⋮2015\)
<=> \(2^x\equiv1\left(mod2015\right)\) (*)
+) With x = 0 then (*) is true
+) With x = 1 then (*) is wrong
+) With x = 2 then (*) is wrong
+) With \(x\ge3\) => \(\left[{}\begin{matrix}x=3k\\x=3k+1\\x=3k+2\end{matrix}\right.\left(k>0\right)\)
-) x = 3k => (*) <=> \(2^{3k}\equiv1\left(mod2015\right)\) <=> \(8^k\equiv1\left(mod2015\right)\) (wrong , because k > 0)
Similar , we can show that with x = 3k + 1 , x = 3k + 2 then (*) is still wrong
So x = 0 is satisfy
Change x = 0 into (*)
=> \(1-2016^{y+1}⋮2015\)
For \(1-2016^{y+1}⋮2015\) then 2016y + 1 = 1
<=> y + 1 = 0
<=> y = -1
Change x = 0 , y = -1 into A , we have A = 0 (satisfy)
Conclude : With x = 0 , y = -1 then A is a square number
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Denote \(\left\{{}\begin{matrix}A=\dfrac{2017^x-2016^{y+1}}{2015}\\B=2017^x-2016^{y+1}\\C=2015\end{matrix}\right.\)
For A is a square number so that \(B⋮C\)
<=> \(2017^x-2016^{y+1}⋮2015\) (1)
<=> 2017x - 2x + 2x - 2016y + 1 + 1 - 1 \(⋮\) 2015
<=> 2x - 1 \(⋮\) 2015
=> \(2^x\equiv1\left(mod2015\right)\) (*)
+) With x = 0 then (*) is true
+) With x = 1 then (*) is false
+) With x = 2 then (*) is false
+) With x \(\ge3\) => \(\left[{}\begin{matrix}x=3k\\x=3k+1\\x=3k+2\end{matrix}\right.\left(k>0\right)\)
Change three cases into (*) , it's still wrong
So , x = 0 is satisfyBecause x = 0 => (1) <=> \(1-2016^{y+1}⋮2015\)
For \(1-2016^{y+1}⋮2015\) then 2016y+1 = 1 => y + 1 = 0 => y = -1
Change x = 0 , y = -1 into A
A = 0 (satisfy the problem)
Conclude : With x = 0 , y = -1 then A is a square number
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Because a,b,c are positive interger numbers so that
a,b,c > 0
If a,b,c are even numbers then \(a,b,c\ge2\)
If a,b,c are odd numbers then \(a,b,c\ge1\)
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2aa + bb = 3cc
+) With c is an odd number
=> 3cc is an odd number => 2aa + bb is an odd number
<*> With a is an even number => 2aa is an even number => b is an odd number
=> \(\left\{{}\begin{matrix}a\ge2\\b\ge1\\c\ge1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2a^a+b^b\ge2.2^2+1^1=7\\3c^c\ge3\end{matrix}\right.\) (removed)
<*> With a is an odd number => 2aa is an even number => b is an odd number
=> \(\left\{{}\begin{matrix}a\ge1\\b\ge1\\c\ge1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2a^a+b^b\ge2.1^1+1^1=3\\3c^c\ge3\end{matrix}\right.\) (satisfy)
+) With c is an even number
=> 3cc is an even number => 2aa + bb is an even number
<*> With a is an even number => 2aa is an even number => bb is an even number
=> \(\left\{{}\begin{matrix}a\ge2\\b\ge2\\c\ge2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2a^a+b^b\ge2.2^2+2^2=12\\3c^c\ge3.2^2=12\end{matrix}\right.\) (satisfy)
<*> With a is an odd number => 2aa is an even number => bb is an even number
=> \(\left\{{}\begin{matrix}a\ge1\\b\ge2\\c\ge2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2a^a+b^b\ge2.1^1+2^2=6\\3c^c\ge3.2^2=12\end{matrix}\right.\) (removed)
From the above analysis, we can see , if a = b = c > 0 then the equation is satisfy .
So P = \(2015^{a-b}+2016^{b-c}+2017^{c-a}=2015^0+2016^0+2017^0=1+1+1=3\)
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Applying inequality AM-GM for two non-negative numbers , we have
\(x+y\ge2\sqrt{xy}\)
=> \(xy\le\dfrac{1}{4}\)
Analyze expression A
=> \(A=\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{x^2+y^2}{x^2y^2}\ge\dfrac{2xy}{x^2y^2}=\dfrac{2}{xy}\ge\dfrac{2}{\dfrac{1}{4}}=8\)
So MinA = 8
Equation occurs when and only when \(x=y=\dfrac{1}{2}\)
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\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
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\(\dfrac{1+2+3+....+100}{2+4+6+.....+100}=\dfrac{\sum\limits^{100}_{x=1}\left(X\right)}{\sum\limits^{50}_{x=1}\left(2X\right)}=\dfrac{101}{51}\)
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Need to prove :<
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
We have :
a3 + b3 + c3 - 3abc
= (a + b)3 - 3ab(a + b) + c3 - 3abc
= [(a + b)3 + c3] - 3ab(a + b + c)
= (a + b + c)[(a + b)2 - (a + b).c + c2]- 3ab(a + b + c)
= (a + b + c)(a2 + 2ab + b2 - ac - bc + c2 - 3ab)
= (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
So :
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
But a + b + c = 0
=> a3 + b3 + c3 - 3abc = 0
=> a3 + b3 + c3 = 3abc
=> a3 + b3 + c3 = 3.11 = 33
So ........
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3n + 2 - 2n + 2 + 3n - 2n
= 3n.9 + 3n - 2n.4 - 2n
= 3n.10 - 2n.5
= 3n . 10 - 2n - 1 . 10
= 10.(3n - 2n - 1) \(⋮10\forall n\)
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No :< , we can do this !
Denote \(t=\dfrac{1}{ab}\) ; \(\dfrac{1}{t}=ab\)
\(t=\dfrac{1}{ab}\ge\dfrac{1}{\left(\dfrac{a+b}{2}\right)^2}=\dfrac{1}{\left(\dfrac{1}{2}\right)^2}=4\)
Predict a = b = \(\dfrac{1}{2}\) then t = 4 , we have :
\(S=\dfrac{1}{t}+t=\left(\dfrac{1}{t}+\dfrac{t}{16}\right)+\dfrac{15t}{16}\ge2.\sqrt{\dfrac{1}{t}.\dfrac{t}{16}}+\dfrac{15.4}{16}=\dfrac{17}{4}\)
So \(Min_S=\dfrac{17}{4}\)
\(\Leftrightarrow a=b=\dfrac{1}{2}\)