Nhận thấy x2 + 1 \(\ge\)1 > 0 \(\forall\)x
=> \(\left(2x^2-3\right)\left(3x^2-\frac{1}{0,12}\right)\left(x^2+1\right)=0\)
<=> \(\orbr{\begin{cases}2x^2-3=0\\3x^2-\frac{1}{0,12}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x^2=3\\3x^2=\frac{1}{0,12}\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=\frac{3}{2}\\x^2=\frac{1}{0,36}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm\sqrt{\frac{3}{2}}\\x=\pm\frac{1}{0,6}\end{cases}}\)
Vậy \(x\in\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}};-\frac{1}{0,6};\frac{1}{0,6}\right\}\)là giá trị cần tìm
\(\left(2x^2-3\right)\left(3x^2-\frac{1}{0,12}\right)\left(x^2+1\right)=0\)
Nhận thấy rằng x2 + 1 ≥ 1 > 0 ∀ x
=> \(\left(2x^2-3\right)\left(3x^2-\frac{1}{0,12}\right)\left(x^2+1\right)=0\)
<=> \(\orbr{\begin{cases}2x^2-3=0\\3x^2-\frac{1}{0,12}=0\end{cases}}\)
+) 2x2 - 3 = 0
<=> 2x2 = 3
<=> x2 = 3/2
<=> x = \(\pm\sqrt{\frac{3}{2}}\)
+) 3x2 - 1/0,12 = 0
<=> 3x2 - 25/3 = 0
<=> 3x2 = 25/3
<=> x2 = 25/9
<=> x = \(\pm\frac{5}{3}\)
Vậy S = { \(\pm\frac{5}{3}\); \(\pm\sqrt{\frac{3}{2}}\))
