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(2x)^-1++1/2x^-1-1
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Your problem has two "+'s" is there supposed to be a number between them?
\[(2x)^{-1} + \frac{ 1 }{ 2x^{-1}} -1 = \frac{ 1 }{ 2x } + 2x - 1\]....now can you do it?
\[\frac{ 1 + 4x^2 + 2x }{ 2x } => (2x +1 )(2x +1)/2x\]
thanks guys
\[\left( 2x \right)^{-1}+\frac{ 1 }{ 2 ~x ^{-1} }-1=\frac{ 1 }{ 2 x }+\frac{ x }{ 2 }-1=\frac{ 1+x^2-2x }{ 2x }\]\[=\frac{ \left( x-1 \right)^2 }{ 2 x }\]
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