Đáp án:
f(x)>0⇒ S=\((-2;-1)∩(-1;3)∩(4;+\infty)\)
f(x)<0⇒ \(S=(-\infty;-2)∩(3;4)\)
f(x)≥0⇒ \(S=[-2;-1)∩(-1;3]∩(4;+\infty)\)
f(x)≤0⇒ \(S=(-\infty;-2]∩[3;4)\)
Giải thích các bước giải:
\(x-\frac{x^{2}-x+6}{-x^{2}+3x+4}=\frac{-x^{3}+3x^{2}+4x-x^{2}+x-6}{-x^{2}+3x+4}\)
=\(\frac{-x^{3}+2x^{2}+5x-6}{-x^{2}+3x+4}=\frac{(-x^{3}+x^{2})+(x^{2}+5x-6)}{(x-4)(x+1)}\)
=\(\frac{-x^{2}(x-1)+(x-1)(x+6)}{(x-4)(x+1)}=\frac{(x-1)(-x^{2}+x+6)}{(x-4)(x+1)}\)
=\(\frac{(x-1)(x-3)(x+2)}{(x-4)(x+1)}\)
Cho x-1=0⇒ x=1
x-3=0⇒ x=3
x+2=0⇒ x=-2
x-4=0⇒ x=4
x+1=0⇒ x=-1
Vậy f(x)>0⇒ S=\((-2;-1)∩(-1;3)∩(4;+\infty)\)
f(x)<0⇒ \(S=(-\infty;-2)∩(3;4)\)
f(x)≥0⇒ \(S=[-2;-1)∩(-1;3]∩(4;+\infty)\)
f(x)≤0⇒ \(S=(-\infty;-2]∩[3;4)\)
