1. Let real numbers $x, y$ satisfy the equation
$$
2 x^{2}+3 y^{2}=4 x \text {. }
$$

Then the minimum value of $x+y$ is ( ).
(A) $1+\frac{\sqrt{15}}{3}$
(B) $1-\frac{\sqrt{15}}{3}$
(C) 0
(D) None of the above
- 1. B.

Equation (1) can be transformed into
$$
(x-1)^{2}+\frac{y^{2}}{\frac{2}{3}}=1 \text {. }
$$

Let $x+y=t$.
When the line (3) is tangent to the ellipse (2), $t$ takes its extreme values.
At this point, $2 x^{2}+3(x-t)^{2}=4 x$, which simplifies to
$$
5 x^{2}-(6 t+4) x+3 t^{2}=0
$$

It has equal roots.
Thus, $\Delta=(6 t+4)^{2}-60 t^{2}=0$
$\Rightarrow t=1 \pm \frac{\sqrt{15}}{3}$.
Therefore, the minimum value of $x+y$ is $1-\frac{\sqrt{15}}{3}$.