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4/x + 4/x^2 /// 3x/16
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4 over x + 4 over x^2 all ov that over 3x over 14
ok. Can you follow this link. http://www.twiddla.com/545962 It is eassy to explain
\[\frac{\frac{\frac{4}{x}}{x^2}}{\frac{3x}{14}}\]
\[a/b/c/d/e/f/g=\frac{a}{b c d f g h}=a*b^{-1}* c^{-1}* . . .*g^{-1} \]\[4*(x+4)^{-1}*(x{}^{\wedge}2)^{-1}*1^{-1}*1^{-1}1*(3x)^{-1}*16^{-1}=\frac{1}{12 x^3 (x+4)}\]
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\[*1^{-1}1* \]should have been\[*1^{-1}*\]
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