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Answers ( 272 )
  • See question detail

    1. 12 = \(6+5+\dfrac{3}{3}\)

    2. 12 = \(3+3+6\)

    3. 12 = \(\sum\limits^6_3\left(X\right)-6\)

    4. 12 = \(\prod\limits^6_5\left(X\right)-3\cdot5-3\)

  • See question detail

    \(4^{50}=2^{100}=2^{10}.2^{10}=\left(...4\right).\left(...4\right)=...6\)

    So ....... 

  • See question detail

    How about using Mincopsky's Inequality ? 

  • See question detail

    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 !

  • See question detail

    +) 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 

  • See question detail

    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 

  • See question detail

    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 = \(a_{2017}=\left(-1\right)^{2017}.\dfrac{2017^2+2017+1}{2017!}=-\dfrac{2017^2+2018}{2017!}\)

  • See question detail

    \(\dfrac{x+b}{x-5}+\dfrac{x+5}{x-b}=2\)

    \(\Rightarrow\dfrac{x+b-\left(x-5\right)}{x-5}+\dfrac{x+5-\left(x-b\right)}{x-b}=0\)

    \(\Rightarrow\dfrac{b+5}{x-5}+\dfrac{b+5}{x-b}=0\)

    \(\Rightarrow\left(b+5\right)\left(\dfrac{1}{x-5}+\dfrac{1}{x-b}\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}b+5=0\\\dfrac{1}{x-5}+\dfrac{1}{x-b}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=-5\\x-b+x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=-5\\x=\dfrac{b+5}{2}\end{matrix}\right.\)

    +) 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 \(\left(\infty;-5\right)\) ;  \(\left(\dfrac{b+5}{2};\infty\right)\)

  • See question detail

    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)

    => \(\Delta MED=\Delta ACB\left(e-a-e\right)\)

  • See question detail

    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 

  • See question detail

    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 

  • See question detail

    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 satisfy 

    Because 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 

  • See question detail

    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 \(a,b,c\ge1\)

  • See question detail

    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 = \(2015^{a-b}+2016^{b-c}+2017^{c-a}=2015^0+2016^0+2017^0=1+1+1=3\)

  • See question detail

    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 \(x=y=\dfrac{1}{2}\)

  • See question detail

    \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)

  • See question detail

    \(\dfrac{1+2+3+....+100}{2+4+6+.....+100}=\dfrac{\sum\limits^{100}_{x=1}\left(X\right)}{\sum\limits^{50}_{x=1}\left(2X\right)}=\dfrac{101}{51}\)

  • See question detail

    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 ........ 

  • See question detail

    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)  \(⋮10\forall n\)

  • See question detail

    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 \(Min_S=\dfrac{17}{4}\) 

    \(\Leftrightarrow a=b=\dfrac{1}{2}\)

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