Question

Please help I can't figure it out

Answer 1

Answer: $\bold{(C)\ \dfrac{x}{4y^2}-\dfrac{5y}{3}-\dfrac{3}{2x}}$

Step-by-step explanation:

Divide each term in the numerator (top) by the entire denominator (bottom) and cross out their common factor(s).

$\dfrac{3x^2}{12xy^2} = \dfrac{3x(x)}{3x(4y^2)}=\boxed{\dfrac{x}{4y^2}}$

$\dfrac{20xy^3}{12xy^2} = \dfrac{4xy^2(5y)}{4xy^2(3)}=\boxed{\dfrac{5y}{3}}$

$\dfrac{18y^2}{12xy^2} = \dfrac{6y^2(3)}{6y^2(2x)}=\boxed{\dfrac{3}{2x}}$

Answer 2

$\frac{3 {x}^{2} \: - \: 20x {y}^{3} \: - \: 18 {y}^{2} }{12x {y}^{2} } \: = \: \frac{3 {x}^{2} }{12x {y}^{2} } \: - \: \frac{20x {y}^{3} }{12x {y}^{2} } \: - \: \frac{18 {y}^{2} }{12x {y}^{2} } \: = \: \frac{x}{4 {y}^{2} } \: - \: \frac{5y}{3} \: - \: \frac{ 3}{2x}$
C.