Toán Lớp 10: Giải bất phương trình: $\frac{x^{2}-3x+2}{(1-|x-2|)x^{2}}$ $\leq$ 0
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Toán Lớp 10: Giải bất phương trình: $\frac{x^{2}-3x+2}{(1-|x-2|)x^{2}}$ $\leq$ 0
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(x^2 – 3 x + 2)/((1 – abs(x – 2)) x^2)<=0
⇔(x^2 – 3 x + 2)/((1 – abs(x – 2)) x^2)<=0
⇔((x – 2) (x – 1))/(x^2 (abs(x – 2) – 1))>=0
⇔-((x – 2) (x – 1))/(x^2 (abs(x – 2) – 1))<=0
⇔(x^2 – 3 x + 2)/(x^2 (abs(x – 2) – 1))>=0
⇔((x – 2) (x – 1))/(abs(x – 2) – 1)>=0
⇔2/(x^2 (1 – abs(x – 2))) – 3/(x (1 – abs(x – 2))) + 1/(1 – abs(x – 2))<=0
⇔2/(x^2 (1 – sqrt((x – 2)^2))) – 3/(x (1 – sqrt((x – 2)^2))) + 1/(1 – sqrt((x – 2)^2))<=0
⇔{(\x>3),(0<x\le \2),(x<0):}
$⇔\begin{cases} (3, ∞)\\(0, 2]\\(-∞, 0) \end{cases}$
$⇔x=2$
Vậy nghiệm của BPT là $x=2$