26 On the $x O y$ coordinate plane, the curve
$$
y=(3 x-1)\left(\sqrt{9 x^{2}-6 x+5}+1\right)+(2 x-3)\left(\sqrt{4 x^{2}-12 x+13}+1\right)
$$

intersects the $x$-axis at the point with coordinates $\qquad$ .
$26\left(\frac{4}{5}, 0\right)$. The x-coordinate of the required intersection point, i.e., the root of the equation $(3 x-1)\left(\sqrt{(3 x-1)^{2}+4}+1\right)=$ $(3-2 x)\left(\sqrt{(3-2 x)^{2}+4}+1\right)$. Consider $f(t)=t\left(\sqrt{t^{2}+4}+1\right)$, it is easy to know that $f(-t)$ $=-f(t), f(t)$ is an odd function. When $t_{1}>t_{2} \geqslant 0$, it is easy to prove that $f\left(t_{1}\right)>f\left(t_{2}\right) . f(t)$ is a strictly increasing function, so over the entire real number set, $f(t)$ is monotonically increasing. From $f(3 x-1)=f(3-2 x)$ we get $3 x-1=3-2 x$, so $x=\frac{4}{5}$.