求满足 $2x(x-3) \ge 3-5x$ 的 $x$ 的取值范围。 求最大整数 $k$ 的值，使得直线 $2y+k=x$ 与曲线 $y=2x+\frac{6}{x}$ 不相交。

1. Express $y$ in terms of $x$ from the linear equation and substitute into the curve equation

From $2y + k = x$, we have $y = \frac{x - k}{2}$.
Substitute this into $y = 2x + \frac{6}{x}$:
$\frac{x - k}{2} = 2x + \frac{6}{x}$

2. Rearrange the equation into a standard quadratic form $ax^2 + bx + c = 0$

Multiply throughout by $2x$:
$x(x - k) = 4x^2 + 12$
$x^2 - kx = 4x^2 + 12$
$3x^2 + kx + 12 = 0$

3. Apply the discriminant condition for no intersection

For the line to not intersect the curve, the quadratic equation must have no real roots, so $b^2 - 4ac < 0$.
$k^2 - 4(3)(12) < 0$
$k^2 - 144 < 0$

4. Solve the inequality for $k$ and find the greatest integer value

$(k - 12)(k + 12) < 0$
$-12 < k < 12$
Since $k$ is an integer, the greatest value of $k$ is $11$.