Chọn phương án D.

\(\begin{aligned}
&\,2^{x^2-5x+6}+2^{1-x^2}=2\cdot2^{6-5x}+1\\
\Leftrightarrow&\,2^{(7-5x)-\left(1-x^2\right)}+2^{1-x^2}=2^{7-5x}+1\\
\Leftrightarrow&\,\dfrac{2^{7-5x}}{2^{1-x^2}}+2^{1-x^2}=2^{7-5x}+1\\
\Leftrightarrow&\,2^{7-5x}+2^{1-x^2}\cdot2^{1-x^2}=2^{1-x^2}\cdot2^{7-5x}+2^{1-x^2}\\
\Leftrightarrow&\,2^{7-5x}+2^{1-x^2}\cdot2^{1-x^2}-2^{1-x^2}\cdot2^{7-5x}-2^{1-x^2}=0\\
\Leftrightarrow&\,2^{7-5x}\left[1-2^{1-x^2}\right]-2^{1-x^2}\left[1-2^{1-x^2}\right]=0\\
\Leftrightarrow&\,\left[2^{7-5x}-2^{1-x^2}\right]\cdot\left[1-2^{1-x^2}\right]=0\\
\Leftrightarrow&\left[\begin{array}{l}2^{7-5x}-2^{1-x^2}\\ 1-2^{1-x^2}\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}2^{7-5x}=2^{1-x^2}\\ 2^0=2^{1-x^2}\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}7-5x=1-x^2\\ 0=1-x^2\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x^2-5x+6=0\\ x^2-1=0\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=2\\ x=3\\ x=1\\ x=-1\end{array}\right.
\end{aligned}\)

Vậy phương trình đã cho có \(4\) nghiệm thực phân biệt.