BTGmoderatorDC wrote:Given that x ≠5, is x>{1/(x-5)^2}
Statement #1: x > 0
Statement #2: x > 10
Source: Magoosh
$$x\,\,\mathop > \limits^? \,\,\frac{1}{{{{\left( {x - 5} \right)}^2}}}\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,x{{\left( {x - 5} \right)}^2}\,\,\mathop > \limits^? \,\,1\,\,\,{\text{with}}\,\,{\text{x}} \ne {\text{5}}\,\,}$$
$$\left( 1 \right)\,\,\,x > 0\left\{ \matrix{
\,{\rm{Take}}\,\,{\rm{x = 1}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr
\,{\rm{Take}}\,\,{\rm{x = 5 + }}{1 \over {10}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\left[ {5.1 \cdot {1 \over {100}} < 1} \right] \hfill \cr} \right.$$
$$\left( 2 \right)\,\,\,x > 10\,\,\, \Rightarrow \,\,\,{\left( {x - 5} \right)^2} > 25\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
