Answer
$(4xy+2x^2-9) \div 4xy=1+\frac{x}{2y}-\frac{9}{4xy}$
Work Step by Step
Divide each term of the dividend by the divisor to have:
$=\dfrac{4xy}{4xy}+\dfrac{2x^2}{4xy}-\dfrac{9}{4xy}
\\=1+\frac{x}{2y}-\frac{9}{4xy}$
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