Question

$\frac{3x}{2 } \times \frac{1}{6x}$
$20x^{2} \times \frac{9}{4x {}^{2} }$
Помогите пожалуйста
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Answer 1

$\frac{3x}{2}* \frac{1}{6x} = \frac{1}{2}*\frac{1}{2}=\frac{1}{4}\\20x^{2} * \frac{9}{4x^{2} }=20*\frac{9}{4}=5*9=45\\(\frac{2a^{2} }{b^{3} })^{3}=\frac{(2a^{2} )^{3} }{(b^{3})^{3} }= \frac{8a^{6} }{b^{9} } \\\frac{x^{2} -16}{8x^{3} } * \frac{4x}{x+4} = \frac{(x-4)(x+4)}{2x^{2} } * \frac{1}{x+4} =\frac{x-4}{2x^{2} } \\\frac{y}{3y^{2}-12 } *(y^{2} -4y+4)=\frac{y}{3(y^{2} -4}) *(y-2)^{2} = \frac{y}{3(y-2)(y+2)} * (y-2)^{2} =\frac{y}{3(y+2)} * (y-2)=\frac{y^{2} -2y}{3(y+2)} =\frac{y^{2}-2y }{3y+6}$