Question

Помогите пожалуйста

Answer 1

Ответ:

а) $\frac{15x^3y^2}{14a^2} : \frac{20x^2y^3}{21a^2b} =\frac{15x^3y^2}{14a^2b^2} * \frac{21a^2b}{20x^2y^3} =\frac{3x}{2b} * \frac{3}{4y} = \frac{9x}{8by}$

б) $\frac{2+y}{2^2-y^2} * \frac{2}{2+y} =\frac{1}{4-y^2} *2=\frac{2}{4-y^2}$

в) $(\frac{3b^2c}{2x} )^4:(3x^2bc^2)^3=\frac{81b^8c^4}{16x^4} : (27x^6b^3c^6)=\frac{81b^8c^4}{16x^4} : (27b^3c^6x^6)=\frac{81b^8c^4}{16x^4} *$

$*\frac{1}{27b^3c^6x^6} =\frac{3b^5}{16x^4} * \frac{1}{c^2x^6} = \frac{3b^5}{16c^2x^{10}}$

г)  $\frac{2x}{(2x-5)^2} * (5-2x)=\frac{2x}{(2x-5)^2} *(-(2x-5))=\frac{2x}{2x-5} * (-1)=-\frac{2x}{2x-5}$

д)  $\frac{x-2y}{4x^2-9y^2} : \frac{x^2-4xy+4y^2}{2x+3y} = \frac{x-2y}{(2x-3y)*(2x+3y)} * \frac{2x+3y}{x^2-4xy+4y^2} =\frac{x-2y}{2x-3y} * \frac{1}{(x-2y)^2} =\frac{1}{2x-3y} *$

$* \frac{1}{x-2y} = \frac{1}{(2x-3y)*(x-2y)} = \frac{1}{2x^2-4xy-3xy+6y^2} = \frac{1}{2x^2-7xy+6y^2}$