Упр.569 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б) (x/(xy+y2) — (x2+y2)/(x3-xy2) + y/(x2-xy)) : (x2-2xy+y2)/(x3+y3) = (x2-xy+y2)/y)x-y);
в) ((2x2y+2xy2)/(7×3+x2y+7xy2+y3) * (7x+y)/(x2-y2) + (x-y)/(x2+y2))*(x2-y2) = x+y;
г) (5/(a2-2a-ax+2x)- 1/(8-8a+2a2) * (20-10a)/(x-2)):25/(x3-8) = (x2+2x+4)/5(a-x);
д) (3a/(9-3x-3a+ax) — 1/(a2-9) : (x-a)/(3a2+9a))* (x2-27)/3a — (x2-3x+9)/(a-x).

Подробный ответ

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)

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