(1) Cheng: 3 (x? -2x-3)& lt； X of 3 (3-3x)? -2x-3 & lt； 3-3x(x+3)(x- 1)& lt； Let f(x)=(x+3)(x- 1)

On the number axis, x=-3 and x= 1 is zero. Obviously, f(-∞)=+∞=f(+∞), so a curve can be drawn according to these characteristics:

ff(-∞)=+∞f(-3)= 0 f( 1)= 0 f(+∞)=+∞

It is known that f (x) < 0 at (-3, 1).

(2) Cheng: [X? -ax-(a+ 1)](x？ +3x+4)& lt； 0 due to x? +3x+4 >0(△& lt； 0)

Just discussing: x? -ax-(a+ 1)& lt； 0, its solution set is (-1, 5), so we know that X =- 1, X = 5 is a=4 substituted by its two roots.