Có bao nhiêu số nguyên $x$ thỏa mãn ${{\log }_{2}}\left( {{x}^{2}}...

Điều kiện $x>0$
Bất phương trình ${{\log }_{2}}\left( {{x}^{2}} \right)+{{\log }_{3}}\left( {{x}^{3}} \right)\ge {{\log }_{2}}x.{{\log }_{3}}x-4$
$\Leftrightarrow 2{{\log }_{2}}x+3{{\log }_{3}}x-{{\log }_{2}}x.{{\log }_{3}}x+4\ge 0$
$\begin{aligned}
& \Leftrightarrow 2{{\log }_{2}}x+3{{\log }_{3}}2.{{\log }_{2}}x-{{\log }_{2}}x.{{\log }_{3}}2.{{\log }_{2}}x+4\ge 0 \\
& \Leftrightarrow {{\log }_{3}}2.{{\left( {{\log }_{2}}x \right)}^{2}}-\left( 2+3{{\log }_{3}}2 \right){{\log }_{2}}x-4\le 0 \\
& \Leftrightarrow \dfrac{2+3{{\log }_{3}}2-\sqrt{4+28{{\log }_{3}}2+9{{\left( {{\log }_{3}}2 \right)}^{2}}}}{2{{\log }_{3}}2}\le {{\log }_{2}}x\le \dfrac{2+3{{\log }_{3}}2+\sqrt{4+28{{\log }_{3}}2+9{{\left( {{\log }_{3}}2 \right)}^{2}}}}{2{{\log }_{3}}2} \\
& \Leftrightarrow {{2}^{\dfrac{2+3{{\log }_{3}}2-\sqrt{4+28{{\log }_{3}}2+9{{\left( {{\log }_{3}}2 \right)}^{2}}}}{2{{\log }_{3}}2}}}\le x\le {{2}^{\dfrac{2+3{{\log }_{3}}2+\sqrt{4+28{{\log }_{3}}2+9{{\left( {{\log }_{3}}2 \right)}^{2}}}}{2{{\log }_{3}}2}}} \\
& \Leftrightarrow 0,53\le x\le 134,08 \\
\end{aligned}$
Do $x$ nguyên nên có $134$ giá trị.