Find the least positive integer satisfying :
{(x+2)² (x-1) x² } / { 5^x (x-2) }   greater than equal to 0.

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Please find below the solution to the asked query:

$$\begin{gathered} \frac{{\left(\mathrm{x}+2\right)}^2\left(\mathrm{x}-1\right){\mathrm{x}}^2}{5^{\mathrm{x}}\left(\mathrm{x}-2\right)}\ge 0 \\ {\left(\mathrm{x}+2\right)}^2,{\mathrm{x}}^2 \mathrm{and} 5^{\mathrm{x}} \mathrm{are} \mathrm{never} \mathrm{negative}, \mathrm{hence} \mathrm{they} \mathrm{can} \mathrm{be} \mathrm{transferred} \mathrm{to} \mathrm{other} \\ \mathrm{end} \mathrm{of} \mathrm{inequality}. \\ \Rightarrow \frac{\left(\mathrm{x}-1\right)}{\left(\mathrm{x}-2\right)}\ge 0 \\ \mathrm{Using} \mathrm{method} \mathrm{of} \mathrm{interval} \\ + + + + + + + 1 - - - - - - - -2 + + ++ + ++ \\ \mathrm{x}\in (-\infty ,1]\cup \left(2,\infty \right) \\ \mathrm{Hence} \mathrm{least} \mathrm{positive} \mathrm{integer} \mathrm{satisfying} \frac{{\left(\mathrm{x}+2\right)}^2\left(\mathrm{x}-1\right){\mathrm{x}}^2}{5^{\mathrm{x}}\left(\mathrm{x}-2\right)}\ge 0 \mathrm{is} \mathrm{x}=1. \end{gathered}$$

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