Question

6. Упростите выражение: (5y ^ 2)/(1 - y ^ 2) / (1 - 1/(1 - y)) b)( x xy-y^ 2 - y x^ 2 -xy ): x^ 2 -y^ 2 8xy​

Answer 1

ответ:

$\displaystyle 1)-\frac{5y}{1+y}\\ \\2)\frac{8}{x-y}$

решение:

$\displaystyle 1)\frac{5y^2}{1-y^2}:(1-\frac{1}{1-y})=\frac{5y^2}{1-y^2}:\frac{1-y-1}{1-y}=\frac{5y^2}{1-y^2}:\frac{-y}{1-y}=\frac{5y^2}{1-y^2}*\frac{1-y}{-y}=\\\frac{5y^2}{1-y^2}*(-\frac{1-y}{y})=-\frac{5y^2}{1-y^2}*\frac{1-y}{y}=-\frac{5y^}{(1-y)(1+y)}*\frac{1-y}{y}=\\\\-\frac{5y}{(1-y)(1+y)}*(1-y)=-\frac{5y}{1+y}$

$\displaystyle 2)(\frac{x}{xy-y^2}-\frac{y}{x^2-xy}):\frac{x^2-y^2}{8xy}=(\frac{x}{y(x-y)}-\frac{y}{x^2-xy}):\frac{x^2-y^2}{8xy}=\\\\(\frac{x}{y(x-y)}-\frac{y}{x(x-y)}):\frac{x^2-y^2}{8xy}=(\frac{x}{y(x-y)}-\frac{y}{x(x-y)})*\frac{8xy}{x^2-y^2}=\\\\\frac{x^2-y^2}{xy(x-y)}*\frac{8xy}{x^2-y^2}=\frac{x^2-y^2}{xy(x-y)}*\frac{8xy}{(x-y)(x+y)}=\frac{(x-y)(x+y)}{xy(x-y)}*\frac{8xy}{(x-y)(x+y)}=$$\displaystyle\\\\\frac{x+y}{xy}*\frac{8xy}{(x-y)(x+y)}=\frac{1}{xy}*\frac{8xy}{x-y} =\frac{1}{y}*\frac{8y}{x-y}=\frac{8}{x-y}$