Đáp án:a)
\(f(x)>0⇒ S=(-\infty;\frac{-2}{3})∩(2;4)\)
f(x)<0⇒ S=\((\frac{-2}{3};2)∩(4;+\infty)\)
f(x)≥0⇒ S=$(-\infty;\frac{-2}{3}]∩(2;4)$
f(x)≤0⇒ $S=[\frac{-2}{3};2)∩(4;+\infty)$
b)
$f(x)>0⇒ S=(\frac{-1}{13};1)$
$f(x)<0⇒ S=(-\infty;\frac{-1}{13})∩(1;+\infty)$
f(x)≥0⇒$S=[\frac{-1}{13};1)$
f(x)⇒$S=(-\infty;\frac{-1}{13}]∩(1;+\infty)$
Giải thích các bước giải:
a)\(f(x)=\frac{3x+2}{(x-4)(2-x)}\)
Cho \(3x+2=0⇒ x=\frac{-2}{3}\)
\(x-4=0⇒ x=4\)
\(2-x=0⇒ x=2\)
Vậy \(f(x)>0⇒ S=(-\infty;\frac{-2}{3})∩(2;4)\)
f(x)<0⇒ S=\((\frac{-2}{3};2)∩(4;+\infty)\)
f(x)≥0⇒ S=$(-\infty;\frac{-2}{3}]∩(2;4)$
f(x)≤0⇒ $S=[\frac{-2}{3};2)∩(4;+\infty)$
b)$f(x)=\frac{2}{1-x}-\frac{3}{5x+2}=\frac{2(5x+2)-3(1-x)}{(1-x)(5x+2)}=\frac{13x+1}{(1-x)(5x+2)}$
Cho $13x+1=0⇒ x=\frac{-1}{13}$
$1-x=0⇒x=1$
$5x+2=0⇒ x=\frac{-2}{5}$
Vậy $f(x)>0⇒ S=(\frac{-1}{13};1)$
$f(x)<0⇒ S=(-\infty;\frac{-1}{13})∩(1;+\infty)$
f(x)≥0⇒$S=[\frac{-1}{13};1)$
f(x)⇒$S=(-\infty;\frac{-1}{13}]∩(1;+\infty)$
