Lời giải:
Đặt \(2^{x^2}=a; 2^x=b\)
PT tương đương:
\(2.(2^{x^2})^2-9.2^{x^2}.2^x+4.(2^x)^2=0\)
\(\Leftrightarrow 2a^2-9ab+4b^2=0\)
\(\Leftrightarrow (a-4b)(2a-b)=0\)
TH1: \(a-4b=0\Leftrightarrow 2^{x^2}=4.2^x\Leftrightarrow 2^{x^2}=2^{x+2}\)
\(\Leftrightarrow x^2=x+2\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
TH2: \(2a-b=0\Leftrightarrow 2^{x^2+1}=2^x\Leftrightarrow x^2-x+1=0\)
\(\Leftrightarrow (x-\frac{1}{2})^2+\frac{3}{4}=0\) (vô lý)
Vậy \(x\in\left\{-1;2\right\}\)