Given condition, \[x\text{ }+\text{ }1/x\text{ }=\text{ }2.5\]

\[x\text{ }+\text{ }1/x\text{ }=\text{ }5/2\]

Taking LCM on L.H.S, we have

\[\left( x2\text{ }+\text{ }1 \right)/x\text{ }=\text{ }5/2\]

\[2\left( x2\text{ }+\text{ }1 \right)\text{ }=\text{ }5x\]

\[2×2\text{ }+\text{ }2\text{ }=\text{ }5x\]

\[2×2\text{ }\text{ }5x\text{ }+\text{ }2\text{ }=\text{ }0\]

\[2×2\text{ }\text{ }4x\text{ }\text{ }x\text{ }+\text{ }2\text{ }=\text{ }0\]

\[2x\left( x\text{ }\text{ }2 \right)\text{ }-\text{ }1\left( x\text{ }\text{ }2 \right)\text{ }=\text{ }0\]

\[\left( 2x\text{ }\text{ }1 \right)\left( x\text{ }\text{ }2 \right)\text{ }=\text{ }0\]

Thus, \[2x\text{ }\text{ }1\text{ }=\text{ }0\] or \[x\text{ }\text{ }2\text{ }=\text{ }0\]

Henceforth,

\[x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\] or \[x\text{ }=\text{ }2\]