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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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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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