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\[\frac{{x}^{2}+4}{3{x}^{3}+4{x}^{2}-4x}\]
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"(x+1)/2+4=7"
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(x^2+4)/(3x^3+4x^2-4x)
(x^2+4) / (3x^3+4x^ 2-4x)
\frac{{x}^{2}+4}{{3x}^{3}+{4x}^{2}-4x}
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