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Solve by using sign charts and express the solutions in interval form.

$$\text{(a)} \frac{(x+2)(x-3)^{2}}{x^{2}+x-2} \geq 0$$ $$\text{(b)} \frac{1}{x-1} < \frac{2}{x}$$

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Is there a certain part you're having trouble with? Typically, full answers are not provided for homework questions, just helpful hints when you get stuck. It would be to your advantage to show any work you've done thus far, or explain why you're having difficulty. –  user80696 Jun 4 '13 at 21:33
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Here is a full solution for $(b)$

$$ \frac{1}{x-1} < \frac{2}{x} \implies \frac{2}{x} - \frac{1}{x-1}>0 \implies \frac{x-2}{x(x-1)}>0 $$

$$ \implies \left\{x-2>0\quad \cap \quad x(x-1)>0 \right\} \cup \left\{x-2<0\quad \cap \quad x(x-1)<0 \right\} $$

$$ \implies \left\{x-2>0\, \cap\,(x-1)>0 \right\}\cup \left\{x-2<0\,\cap\left\{( x<0 \cap x-1>0 ) \cup ( x>0 \cap x-1<0 )\right\}\right\} $$

$$ \implies \left\{x-2 >0 \right\} \cup \left\{ x-2<0 \,\cap \, 0<x<1 \right\} $$

$$ \implies \left\{x-2 >0 \right\} \cup \left\{ 0<x<1 \right\} $$

$$ \implies (2,\infty) \cup ( 0,1 ) $$

Note: $\cup$ stands for union while $\cap$ stands for intersection.

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