1. Equation
$$
\begin{array}{l}
\left(x^{2}+2011 x-2012\right)^{2}+\left(2 x^{2}-2015 x+2014\right)^{2} \\
=\left(3 x^{2}-4 x+2\right)^{2}
\end{array}
$$

The sum of all real roots of the equation is ( ).
(A) $\frac{2007}{2}$
(B) $-\frac{2007}{2}$
(C) 4
(D) -4
-、1. B.
Notice,
$$
\begin{array}{l}
\left(x^{2}+2011 x-2012\right)+\left(2 x^{2}-2015 x+2014\right) \\
=3 x^{2}-4 x+2 \text {. } \\
\text { Let } x^{2}+2011 x-2012=y, \\
2 x^{2}-2015 x+2014=z .
\end{array}
$$

Then $3 x^{2}-4 x+2=y+z$.
So $y^{2}+z^{2}=(y+z)^{2}$
$$
\begin{array}{l}
\Rightarrow y=0 \text { or } z=0 \\
\Rightarrow x^{2}+2011 x-2012=0
\end{array}
$$

or $2 x^{2}-2015 x+2014=0$.
Therefore, the sum of all real roots of the original equation is
$$
-2011+\frac{2015}{2}=-\frac{2007}{2} .
$$