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(4y^2-3y-5)/(2y^2+2y-2)

Guest Sep 18, 2012

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1⁺⁰ Answers

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what you think you can do is (4ψ^2-3ψ-5)/2ψ^2+2ψ-2)
=[2ψ(2ψ-1)-ψ-5]/[2ψ(ψ+1)-2]

Hellas Sep 30, 2012

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