397. Solve the equation:

$$
\sqrt{9 x^{2}-12 x+11}+\sqrt{5 x^{2}-8 x+10}=2 x-1
$$
$\triangle$ Multiply both sides of the equation by the conjugate of its left part

$$
\sqrt{9 x^{2}-12 x+11}+\sqrt{5 x^{2}-8 x+10}
$$

Since it does not turn into zero, we will obtain an equation equivalent to the given one. We will have:

$$
\begin{aligned}
& \left(9 x^{2}-12 x+11\right)-\left(5 x^{2}-8 x+10\right)= \\
& =(2 x-1)\left(\sqrt{9 x^{2}-12 x+11}+\sqrt{5 x^{2}-8 x+10}\right) \\
& (2 x-1)^{2}=(2 x-1)\left(\sqrt{9 x^{2}-12 x+11}+\sqrt{5 x^{2}-8 x+10}\right) \\
& (2 x-1)\left(2 x-1-\sqrt{9 x^{2}-12 x+11}-\sqrt{5 x^{2}-8 x+10}\right)=0
\end{aligned}
$$

Consider two cases.

1) Let $2 x-1=0$. Then $x=\frac{1}{2}$.
2) Let the expression in the second parentheses of the left part of the equation be equal to zero. We obtain a system of two equations with one unknown:

$$
\left\{\begin{array}{l}
\sqrt{9 x^{2}-12 x+11}+\sqrt{5 x^{2}-8 x+10}=2 x-1 \\
\sqrt{9 x^{2}-12 x+11}-\sqrt{5 x^{2}-8 x+10}=2 x-1
\end{array}\right.
$$

From here

$$
\sqrt{5 x^{2}-8 x+10}=0, \quad 5 x^{2}-8 x+10=0
$$

But the last equation has no real roots.

Answer: $1 / 2$.