F (x) =-3 * x 2+2x-k+ 1, because the quadratic coefficient is negative, the opening is downward. As long as the vertex of F (x) =-3 * x 2+2x-k+ 1 is less than 0, the function will not intersect with the x axis, and the vertex = (. 0, solving k & gt4/3.

Method 2: f (x) =-3 * x 2+2x-k+ 1, because the quadratic coefficient is negative, so the opening is downward, and the function does not intersect with the x axis, that is, -3 * x 2+2x-k+ 1 = 0 has no solution, that is, △ = b 2. 0 ,2^2-4*(-3)*(-k+ 1)<； 0, solving k & gt4/3.