依题意，令lgxx2−7x+12≥0，即lgxx2−7x+12≥lg1．于是有  xx2−7x+12≥1⇒xx2−7x+12−1≥0⇒x2−8x−12x2−7x+12≤0⇒(x−2)(x−6)(x−3)(x−4)≤0⇒(x−2)(x−6)(x−3)(x−4)≤0(x−3)(x−4)≠0⇒x∈[...