Упр.16.3 ГДЗ Мордкович Семенов 8 класс (Алгебра)

1. Упростите выражение:
   
   а) $$\left(\frac{x}{y^2-xy}+\frac{y}{x^2-xy}\right)\cdot\frac{xy}{x+y}$$;
   
   б) $$\left(\frac{bc}{b^2-c^2}+\frac{c}{2c-2b}\right)\cdot\frac{b+c}{5c}$$;
   
   в) $$\frac{y^2-49}{y+3}\cdot\frac{1}{y^2+7y}-\frac{y+7}{y^2-3y}$$;
   
   г) $$\frac{4a+x}{a^2x-ax^2}+\frac{x-a}{ax}:\frac{a^2-x^2}{4a-x}$$.

1. а)

   $$\left(\frac{x}{y^2-xy}+\frac{y}{x^2-xy}\right)\cdot \frac{xy}{x+y}$$
   $$=\left(\frac{x}{y(y-x)}+\frac{y}{x(x-y)}\right)\cdot \frac{xy}{x+y}$$
   $$=\left(\frac{x^2-y^2}{xy(y-x)(x-y)}\right)\cdot \frac{xy}{x+y}$$
   $$=\left(\frac{(x-y)(x+y)}{-xy(x-y)}\right)\cdot \frac{xy}{x+y}=-\frac{x+y}{xy}\cdot \frac{xy}{x+y}=-1.$$

2. б)

   $$\left(\frac{bc}{b^2-c^2}+\frac{c}{2c-2b}\right)\cdot \frac{b+c}{5c}$$
   $$=\left(\frac{bc}{(b-c)(b+c)}+\frac{c}{2(c-b)}\right)\cdot \frac{b+c}{5c}$$
   $$=\left(\frac{2bc-c(b+c)}{2(b-c)(b+c)}\right)\cdot \frac{b+c}{5c}$$
   $$=\frac{bc-c^2}{2(b-c)(b+c)}\cdot \frac{b+c}{5c}$$
   $$=\frac{c(b-c)}{2(b-c)(b+c)}\cdot \frac{b+c}{5c}=\frac{1}{10}.$$

3. в)

   $$\frac{y^2-49}{y+3}\cdot \frac{1}{y^2+7y}-\frac{y+7}{y^2-3y}$$
   $$=\frac{(y-7)(y+7)}{y+3}\cdot \frac{1}{y(y+7)}-\frac{y+7}{y(y-3)}$$
   $$=\frac{y-7}{y(y+3)}-\frac{y+7}{y(y-3)}$$
   $$=\frac{(y-7)(y-3)-(y+7)(y+3)}{y(y-3)(y+3)}$$
   $$=\frac{y^2-10y+21-y^2-10y-21}{y(y-3)(y+3)}$$
   $$=\frac{-20y}{y(y-3)(y+3)}=-\frac{20}{y^2-9}=\frac{20}{9-y^2}.$$

4. г)

   $$\frac{4a+x}{a^2x-ax^2}+\frac{x-a}{ax}:\frac{a^2-x^2}{4a-x}$$
   $$=\frac{4a+x}{ax(a-x)}+\frac{x-a}{ax}\cdot \frac{4a-x}{a^2-x^2}$$
   $$=\frac{4a+x}{ax(a-x)}+\frac{x-a}{ax}\cdot \frac{4a-x}{(a-x)(a+x)}$$
   $$=\frac{4a+x}{ax(a-x)}-\frac{4a-x}{ax(a+x)}$$
   $$=\frac{(4a+x)(a+x)-(4a-x)(a-x)}{ax(a-x)(a+x)}$$
   $$=\frac{4a^2+5ax+x^2-(4a^2-5ax+x^2)}{ax(a-x)(a+x)}$$
   $$=\frac{10ax}{ax(a-x)(a+x)}=\frac{10}{a^2-x^2}.$$

Ответ

а) $$-1$$; б) $$\frac{1}{10}$$; в) $$\frac{20}{9-y^2}$$; г) $$\frac{10}{a^2-x^2}$$.