Упр.1.91 ГДЗ Дорофеев Суворова 8 класс (Алгебра)

а) xy/(x-y)•(1/y^2 -1/x^2 );
б) (mn^2)/(n^2-m^2 )•(2/m-2/n);
в) (a-(6a-4)/(a+2))•(a+2)/(a^2-2a);
г) (u/(u-v)-u/(u+v))•(u^2+uv)/2v;
д) ((c-d)/d+2c/(c-d)) :(c^2+d^2)/(c-d);
е) ((a+b)/a-(a+b)/b) :(a+b)/(a^2 b^2 ).

1. а)

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

2. б)

   $$\frac{mn^2}{n^2-m^2}\cdot\left(\frac{2}{m}-\frac{2}{n}\right) =\frac{mn^2}{(n-m)(n+m)}\cdot\frac{2n-2m}{mn} =\frac{mn^2\cdot 2(n-m)}{(n-m)(n+m)\cdot mn} =\frac{2n}{n+m}.$$

3. в)

   $$\left(a-\frac{6a-4}{a+2}\right)\cdot\frac{a+2}{a^2-2a} =\frac{a(a+2)-6a+4}{a+2}\cdot\frac{a+2}{a^2-2a} =\frac{a^2+2a-6a+4}{a+2}\cdot\frac{a+2}{a^2-2a}$$
   $$=\frac{a^2-4a+4}{a+2}\cdot\frac{a+2}{a^2-2a} =\frac{(a-2)^2}{a(a-2)} =\frac{a-2}{a}.$$

4. г)

   $$\left(\frac{u}{u-v}-\frac{u}{u+v}\right)\cdot\frac{u^2+uv}{2v} =\frac{u(u+v)-u(u-v)}{(u-v)(u+v)}\cdot\frac{u^2+uv}{2v}$$
   $$=\frac{u^2+uv-u^2+uv}{(u-v)(u+v)}\cdot\frac{u^2+uv}{2v} =\frac{2uv}{(u-v)(u+v)}\cdot\frac{u(u+v)}{2v} =\frac{u^2}{u-v}.$$

5. д)

   $$\left(\frac{c-d}{d}+\frac{2c}{c-d}\right):\frac{c^2+d^2}{c-d} =\frac{(c-d)^2+2cd}{d(c-d)}\cdot\frac{c-d}{c^2+d^2}$$
   $$=\frac{c^2-2cd+d^2+2cd}{d(c-d)}\cdot\frac{c-d}{c^2+d^2} =\frac{c^2+d^2}{d(c-d)}\cdot\frac{c-d}{c^2+d^2} =\frac{1}{d}.$$

6. е)

   $$\left(\frac{a+b}{a}-\frac{a+b}{b}\right):\frac{a+b}{a^2b^2} =\frac{b(a+b)-a(a+b)}{ab}\cdot\frac{a^2b^2}{a+b}$$
   $$=\frac{(b-a)(a+b)}{ab}\cdot\frac{a^2b^2}{a+b} =\frac{(b-a)a^2b^2}{ab} =ab(b-a).$$

Ответ

а) $$\frac{x+y}{xy}$$; б) $$\frac{2n}{n+m}$$; в) $$\frac{a-2}{a}$$; г) $$\frac{u^2}{u-v}$$; д) $$\frac{1}{d}$$; е) $$ab(b-a)$$.