Sửa lại đề :
\(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)< 0\)
\(\Leftrightarrow\left[\left(x^2-1\right)\left(x^2-10\right)\right].\left[\left(x^2-4\right)\left(x^2-7\right)\right]< 0\)
\(\Leftrightarrow\left(x^4-11x^2+10\right)\left(x^4-11x^2+28\right)< 0\)
\(\Leftrightarrow x^4-11x^2+10,x^2-11x^2+28\) là 2 số trái dấu .
Mà \(x^4-11x^2+10< x^4-11x^2+28\)
\(\Leftrightarrow\hept{\begin{cases}x^4-11x^2+10< 0\\x^4-11x^2+28>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x^2-\frac{11}{2}\right)^2-\frac{81}{4}< 0\\\left(x^2-\frac{11}{2}\right)^2-\frac{9}{4}>0\end{cases}}\)
\(\Leftrightarrow\frac{9}{4}< \left(x^2-\frac{11}{12}\right)^2< \frac{81}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}< x^2-\frac{11}{2}< \frac{9}{2}\\-\frac{3}{2}>x^2-\frac{11}{2}>-\frac{9}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}7< x^2< 10\\4>x^2>1\end{cases}}\)
Vì \(x\in Z\Leftrightarrow x^2\in Z\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy \(x=3;-3\)
Chúc bạn học tốt !!!
