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