interger

We have: | x-1 |+|x-3|+|x-5|+|x-7|=|x-1|+|x-3|+|5-x|+|7-x|

\(\ge\)|x-1+x-3+5-x+7-x|=|8|=8

Mark ''='' occurs when \(\left(x-1\right)\left(x-3\right)\left(5-x\right)\left(7-x\right)\ge0\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)\ge0\)

Case 1 :x-1 , x-3 , x-5 , x-7 \(\ge0\) so x=7 because if x>7 so (x-1)(x-3)(x-5)(x-7)>0(not satisfy)

Case 2 :x-1 , x-3\(\ge\)0 ; x-5 , x-7\(\le0\)\(\Rightarrow x-3\ge0;x-5\le0\)\(\Rightarrow5\ge x\ge3\)

Case 3 :x-1 , x-5 \(\ge\)0 ; x-3 , x-7\(\le\)0\(\Rightarrow x-5\ge0;x-3\le0\)\(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\)(not satisfy)

Case 4 :x-1 , x-7 \(\ge0\) ; x-3 , x-5\(\le\)0\(\Rightarrow\left\{{}\begin{matrix}x\ge7\\x\le3\end{matrix}\right.\)(not satisfy)

Case 5 :x-1 , x-3 , x-5 , x-7\(\le\)0\(\Rightarrow x\le1\)

Case 6 :x-1 , x-3\(\le0\) ; x-5 , x-7\(\ge\)0\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge7\end{matrix}\right.\)(not satisfy)

Case 7 :x-1 , x-5\(\le\)0 ; x-3 , x-7\(\ge\)0\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge7\end{matrix}\right.\)(not satisfy)

Case 8 :x-1 , x-7\(\le0\) ; x-3 , x-5\(\ge0\)\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge5\end{matrix}\right.\)(not satisfy)

Retry x=7 we have |7-1|+|7-3|+|7-5|+|7-7|=12(not satisfy)

So max_x=5

We have: | x-1 |+|x-3|+|x-5|+|x-7|=|x-1|+|x-3|+|5-x|+|7-x|

≥

|x-1+x-3+5-x+7-x|=|8|=8

Mark ''='' occurs when (x−1)(x−3)(5−x)(7−x)≥0

⇒(x−1)(x−3)(x−5)(x−7)≥0

Case 1 :x-1 , x-3 , x-5 , x-7 ≥0

 so x=7 because if x>7 so (x-1)(x-3)(x-5)(x-7)>0(not satisfy)

Case 2 :x-1 , x-3≥

0 ; x-5 , x-7≤0⇒x−3≥0;x−5≤0⇒5≥x≥3

Case 3 :x-1 , x-5 ≥

0 ; x-3 , x-7≤0⇒x−5≥0;x−3≤0⇒{x≥5x≤3

(not satisfy)

Case 4 :x-1 , x-7 ≥0

 ; x-3 , x-5≤0⇒{x≥7x≤3

(not satisfy)

Case 5 :x-1 , x-3 , x-5 , x-7≤

0⇒x≤1

Case 6 :x-1 , x-3≤0

 ; x-5 , x-7≥0⇒{x≤1x≥7

(not satisfy)

Case 7 :x-1 , x-5≤

0 ; x-3 , x-7≥0⇒{x≤1x≥7

(not satisfy)

Case 8 :x-1 , x-7≤0

 ; x-3 , x-5≥0⇒{x≤1x≥5

(not satisfy)

Retry x=7 we have |7-1|+|7-3|+|7-5|+|7-7|=12(not satisfy)

So maxx=5

Các bạn lập bảng xét dấu: như vậy chỉ cần xét 5 trường hợp thôi:

1)  \(x\le1\) thì các biểu thức trong trị tuyệt đối đều âm ta có  VT=(1-x)+(3-x)+(5-x)+(7-x)=8

ta giải tìm x

Tương tự cho 4 trường hợp còn lại. 

For \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)< 0\)

=> \(x^2-1;x^2-4;x^2-7;x^2-10\) have a positive and 3 negative or a nagative and 3 positive

* If have a positive and 3 negative:

=> x² - 1 > 0 > x² - 4 > x² - 7 > x² - 10

=> x² - 1 > 0 and x² - 4 < 0

=> 1 < x² < 4 => x² can equal 2 or 3.

 But x is integer so x² can't equal 2 or 3

* If have a negative and 3 positive:

=> x² - 1 > x² - 4 > x² - 7 > 0 > x² - 10

=> x² - 7 > 0 and x² - 10 < 0

=> 7 < x² < 10

=> x² = 9 because x is integer

=> x = \(\pm3\)

But x is the positive integer => x = 3

Equivalent disequations:\(\left(x^4-11x^2+10\right)\left(x^4-11x^2+28\right)< 0\)

Put \(x^4-11x^2+10=t\)

We have:t(t+18)<0

TH1:\(\left\{{}\begin{matrix}t>0\\t+18< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}t>0\\t< -18< 0\end{matrix}\right.\) (loại)

TH2:\(\left\{{}\begin{matrix}t< 0\\t+18>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}t< 0\\t>-18\end{matrix}\right.\)\(\Rightarrow t\in\left\{-17,-16,...........,-1\right\}\)

It's easy here, you try it yourself......

For \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-1\right)< 0\)

=> In this four numbers, have:

\(\left[{}\begin{matrix}\text{a positive number and 3 negative numbers}\\\text{a negative number and 3 positive numbers}\end{matrix}\right.\)

* If have a positive number and 3 negative numbers:

=> x²-1 > 0 > x²-4 > x²-7 > x²-10

=> x² - 1 > 0 and x² < 4

=> 1 < x² < 4 => No number is satisfy

* If have a negative number and 3 positive numbers:

=> x²-1 > x²-4 > x²-7 > 0 > x²-10

=> x²-7 > 0 and x²-10 < 0

=> 7 < x² < 10

=> x² = 9 because x is a positive integer <=> x² is a positive integer too.

=> x = 3 because x is a positive integer

Without loss of generality, we may assume gcd(a,b,c) = 1.

(otherwise, if d=gcd(a,b,c) the for a'=a/d, b'=b/d, c'=c/d, the equation still holds for a', b', c' and a'b'c' is still a cube if only if abc is a cube).

We multiply equation by abc, we have:

  \(a^2c+b^2a+c^2b=3abc\)(*)

if \(abc=\pm1\), the problem is solved.

Otherwise, let p be a prime divisor of abc. Since gcd(a,b,c)=1, the (*) implies that p divides exactly two of a, b,c. By symetry, we may assume p divides a, b but not c. Suppose that the lagest powers of p dividing a, b are m, n, respecively.

If n < 2m, then \(n+1\le2m\)  and \(p^{n+1}\)| \(a^2c,b^2c,3abc\). Hence \(p^{n+1}\)|\(c^2b\), forcing \(\)\(p\)|\(c\) (a contradiction). If n > 2m, then \(n\ge2m+1\) and \(p^{2m+1}\)|\(c^2b,b^2a,3abc\). Hence \(p^{2m+1}\)|\(a^2c\), forcing \(p\)|\(c\) (a contradicton). Therefore n = 2m and \(abc=\Pi p^{3m}\), \(p\)|\(abc\), is a cube.

We have:

\(\left(x+y+z\right)^3-\left(x^3+y^3+z^3\right)=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Rightarrow8=\left(3-z\right)\left(3-x\right)\left(3-y\right)\)   (1)

Since \(\left(3-x\right)+\left(3-y\right)+\left(3-z\right)=9-\left(x+y+z\right)=9-3=6\)  (2)

From (1) and (2) we infer \(\left(x,y,z\right)\) would be (1,1,1), (-5,4,4), (4,-5,4), (4,4,-5).

We have:

(x+y+z)3−(x3+y3+z3)=3(x+y)(y+z)(z+x)

⇒8=(3−z)(3−x)(3−y)

   (1)

Since (3−x)+(3−y)+(3−z)=9−(x+y+z)=9−3=6

  (2)

From (1) and (2) we infer (x,y,z)

 would be (1,1,1), (-5,4,4), (4,-5,4), (4,4,-5).