Solving 1x−2+1x−1&gt;1x

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Solving 1x−2+1x−1&amp;gt;1x

Cailyn Hanson · Accepted Answer

Step 1Actually,1x−2+1x−1−1x=x2−2x(x−1)(x−2).Besides:x2−2&amp;gt;0 if and only ifx∈(−∞,−2)∪(2,∞)x2−2&amp;lt;0 if and only if x∈(−2,2)x(x−1)(x−2)&amp;gt;0 if and only if x∈(0,1)∪(2,∞)x(x−1)(x−2)&amp;lt;0 if and only if x∈(−∞,0)∪(1,2)

Yaritza Phillips · Accepted Answer

Step 1x∈{0,1,2}1x−2+1x−1&amp;gt;1xIf&amp;nbsp;x&amp;gt;2, then each terms are positive, then clearly it is true.Now, we focus on&amp;nbsp;x&amp;lt;2, that is&amp;nbsp;x−2&amp;lt;01x−1−1x&amp;gt;12−x2−xx(x−1)&amp;gt;12−x−x(x−1)x(x−1)&amp;gt;02−x2x(x−1)&amp;gt;0If&amp;nbsp;2−x2&amp;gt;0, then&amp;nbsp;x(x−1)&amp;gt;0, that is&amp;nbsp;−2&amp;lt;x&amp;lt;2&amp;nbsp;and (x&amp;lt;0&amp;nbsp;or&amp;nbsp;x&amp;gt;1). That is we obtain&amp;nbsp;(−2,0)∪(1,2)If&amp;nbsp;2−x2&amp;lt;0, then&amp;nbsp;x(x−1)&amp;lt;0, that is&amp;nbsp;|x|&amp;gt;2&amp;nbsp;and&amp;nbsp;0≤x≤1, of which no such value exists.Hence in summary,&amp;nbsp;(−2,0)∪(1,2)∪(2,∞)

Solving \(\displaystyle{\frac{{1}}{{{x}-{2}}}}+{\frac{{1}}{{{x}-{1}}}}{>}{\frac{{1}}{{x}}}\)

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