Упр.569 ГДЗ Никольский Потапов 7 класс (Алгебра)

а)

$$\frac{a^2+b^2}{ab}\left(\frac{6a+b}{a^2-b^2}:\frac{6a^3+b^3+a^2b+6ab^2}{2ab^2-2a^2b}+\frac{a+b}{a^2+b^2}\right)$$
$$\frac{6a^3+b^3+a^2b+6ab^2}{2ab^2-2a^2b} =\frac{(6a+b)(a^2+b^2)}{2ab(b-a)}$$
$$\frac{6a+b}{a^2-b^2}:\frac{(6a+b)(a^2+b^2)}{2ab(b-a)} =\frac{6a+b}{(a-b)(a+b)}\cdot\frac{2ab(b-a)}{(6a+b)(a^2+b^2)} =-\frac{2ab}{(a+b)(a^2+b^2)}$$
$$-\frac{2ab}{(a+b)(a^2+b^2)}+\frac{a+b}{a^2+b^2} =\frac{-2ab+(a+b)^2}{(a+b)(a^2+b^2)} =\frac{a^2+b^2}{(a+b)(a^2+b^2)} =\frac1{a+b}$$
$$\frac{a^2+b^2}{ab}\cdot\frac1{a+b} =\frac{a^2+b^2}{ab(a+b)}.$$

б)

$$\left(\frac{x}{xy+y^2}-\frac{x^2+y^2}{x^3-xy^2}+\frac{y}{x^2-xy}\right):\frac{x^2-2xy+y^2}{x^3+y^3}$$
$$\frac{x}{y(x+y)}-\frac{x^2+y^2}{x(x-y)(x+y)}+\frac{y}{x(x-y)} =\frac{x^2-2xy+y^2}{xy(x+y)(x-y)} =\frac{(x-y)^2}{xy(x+y)(x-y)} =\frac{x-y}{y(x+y)}$$
$$\frac{x-y}{y(x+y)}:\frac{(x-y)^2}{x^3+y^3} =\frac{x-y}{y(x+y)}\cdot\frac{x^3+y^3}{(x-y)^2}$$
$$x^3+y^3=(x+y)(x^2-xy+y^2)$$
$$\frac{x-y}{y(x+y)}\cdot\frac{(x+y)(x^2-xy+y^2)}{(x-y)^2} =\frac{x^2-xy+y^2}{y(x-y)}.$$

в)

$$\left(\frac{2x^2y+2xy^2}{7x^3+x^2y+7xy^2+y^3}\cdot\frac{7x+y}{x^2-y^2}+\frac{x-y}{x^2+y^2}\right)\cdot(x^2-y^2)$$
$$\frac{2x^2y+2xy^2}{7x^3+x^2y+7xy^2+y^3} =\frac{2xy(x+y)}{(7x+y)(x^2+y^2)}$$
$$\frac{2xy(x+y)}{(7x+y)(x^2+y^2)}\cdot\frac{7x+y}{x^2-y^2} =\frac{2xy}{(x^2+y^2)(x-y)(x+y)}\cdot(x+y) =\frac{2xy}{(x^2+y^2)(x-y)}$$
$$\frac{2xy}{(x^2+y^2)(x-y)}+\frac{x-y}{x^2+y^2} =\frac{2xy+(x-y)^2}{(x^2+y^2)(x-y)} =\frac{x^2+y^2}{(x^2+y^2)(x-y)} =\frac1{x-y}$$
$$\frac1{x-y}\cdot(x^2-y^2)=\frac{(x-y)(x+y)}{x-y}=x+y.$$

г)

$$\left(\frac{5}{a^2-2a-ax+2x}-\frac{1}{8-8a+2a^2}\cdot\frac{20-10a}{x-2}\right):\frac{25}{x^3-8}$$
$$8-8a+2a^2=2(a-2)^2,\qquad 20-10a=10(2-a)$$
$$\frac{1}{8-8a+2a^2}\cdot\frac{20-10a}{x-2} =\frac{1}{2(a-2)^2}\cdot\frac{10(2-a)}{x-2} =\frac{5}{(2-a)(x-2)}$$
$$a^2-2a-ax+2x=(a-2)(a-x)$$
$$\frac{5}{(a-2)(a-x)}-\frac{5}{(2-a)(x-2)} =\frac{5}{(a-2)(a-x)}+\frac{5}{(a-2)(x-2)}$$
$$=\frac{5(x-2)+5(a-x)}{(a-2)(a-x)(x-2)} =\frac{5(a-2)}{(a-2)(a-x)(x-2)} =\frac{5}{(a-x)(x-2)}$$
$$x^3-8=(x-2)(x^2+2x+4)$$
$$\frac{5}{(a-x)(x-2)}:\frac{25}{x^3-8} =\frac{5}{(a-x)(x-2)}\cdot\frac{(x-2)(x^2+2x+4)}{25} =\frac{x^2+2x+4}{5(a-x)}.$$

д)

$$\left(\frac{3a}{9-3x-3a+ax}-\frac{1}{a^2-9}:\frac{x-a}{3a^2+9a}\right)\cdot\frac{x^3-27}{3a}$$
$$\frac{1}{a^2-9}:\frac{x-a}{3a^2+9a} =\frac{1}{(a-3)(a+3)}\cdot\frac{3a(a+3)}{x-a} =\frac{3a}{(a-3)(x-a)}$$
$$9-3x-3a+ax=(3-x)(3-a)$$
$$\frac{3a}{(3-x)(3-a)}-\frac{3a}{(a-3)(x-a)} =\frac{3a}{(3-x)(3-a)}+\frac{3a}{(3-a)(x-a)}$$
$$=\frac{3a(x-a)+3a(3-x)}{(3-x)(3-a)(x-a)} =\frac{3a(3-a)}{(3-x)(3-a)(x-a)} =\frac{3a}{(3-x)(x-a)}$$
$$x^3-27=(x-3)(x^2+3x+9)$$
$$\frac{3a}{(3-x)(x-a)}\cdot\frac{x^3-27}{3a} =\frac{3a}{(3-x)(x-a)}\cdot\frac{(x-3)(x^2+3x+9)}{3a} =\frac{x^2+3x+9}{a-x}.$$

Ответ

а) $$\frac{a^2+b^2}{ab(a+b)}$$; б) $$\frac{x^2-xy+y^2}{y(x-y)}$$; в) $$x+y$$; г) $$\frac{x^2+2x+4}{5(a-x)}$$; д) $$\frac{x^2+3x+9}{a-x}$$.