Question

ПОМОГИТЕ СРОЧНО

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Answer 1

Ответ:

Пошаговое объяснение:

a)

$(\frac{x}{x-1} -1)*\frac{xy-y}{y} =(\frac{x}{x-1} -1) * \frac{y(x-1)}{x} =\frac{x}{x-1} *\frac{y(x-1)}{x} -\frac{y(x-1)}{x}=y-\frac{y(x-1)}{x}=\frac{xy}{x} -\frac{y(x-1)}{x}=\frac{xy-xy+y}{x} =\frac{y}{x}$

б)

$\frac{a}{b} -\frac{a^{2}-b^{2} }{b^{2} } :\frac{a+b}{b} =\frac{a}{b} -\frac{(a+b)(a-b)}{b^{2} } *\frac{b}{a+b} =\frac{a}{b} -\frac{(a-b)}{b} =\frac{a-a+b}{b} =\frac{b}{b} =1$

в)

$(\frac{y+3}{y-3} -\frac{y-3}{y+3}):\frac{12y}{y^{2}+6y+9} =(\frac{y+3}{y-3} -\frac{y-3}{y+3})*\frac{(y+3)^{2}}{12y} =\frac{(y+3)(y+3)-(y-3)(y-3)}{(y+3)(y-3)} *\frac{(y+3)^{2}}{12y} =\frac{(y+3)^{2} -(y-3)^{2} }{(y+3)(y-3)} *\frac{(y+3)^{2}}{12y} =\frac{y^{2}+6y+9-(y^{2}-6y+9)}{(y+3)(y-3)} *\frac{(y+3)^{2}}{12y} =\frac{y^{2}+6y+9-y^{2}+6y-9}{(y+3)(y-3)} *\frac{(y+3)^{2}}{12y} =\frac{12y}{(y+3)(y-3)} *\frac{(y+3)^{2}}{12y} =\frac{y+3}{y-3}$