Упр.35 Вариант 1 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)

Задача

1) ((a-2)/(a+2)-(a+2)/(a-2)) :(12a^2)/(4-a^2);
2) (8x/(x-2)+2x) :(4x+8)/(7x-14);
3) 5a/(a+3)+(a-6)/(3a+9)•135/(6a-a^2);
4) (3m/(m+5)-8m/(m^2+10m+25)) :(3m+7)/(m^2-25)+(5m-25)/(m+5);
5) (y^2/(x^3-xy^2 )+1/(x+y)) :((x-y)/(x^2+xy)-x/(xy+y^2 ));
6) (a/(a-4)-a/(a+4)-(a^2+16)/(16-a^2 )) :(4a+a^2)/(4-a)^2 .

Подробный ответ

$$\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}\right):\frac{12a^2}{4-a^2}$$
$$\frac{(a-2)^2-(a+2)^2}{(a+2)(a-2)}:\frac{12a^2}{4-a^2}$$
$$\frac{a^2-4a+4-a^2-4a-4}{a^2-4}:\frac{12a^2}{4-a^2}$$
$$\frac{-8a}{a^2-4}\cdot\frac{4-a^2}{12a^2}=\frac{-8a\cdot (-(a^2-4))}{(a^2-4)\cdot 12a^2}=\frac{2}{3a}$$
$$\left(\frac{8x}{x-2}+2x\right):\frac{4x+8}{7x-14}$$
$$\frac{8x+2x(x-2)}{x-2}:\frac{4x+8}{7x-14}$$
$$\frac{2x^2+4x}{x-2}\cdot\frac{7(x-2)}{4(x+2)}=\frac{2x(x+2)}{x-2}\cdot\frac{7(x-2)}{4(x+2)}=\frac{7x}{2}$$
$$\frac{5a}{a+3}+\frac{a-6}{3a+9}\cdot\frac{135}{6a-a^2}$$
$$\frac{5a}{a+3}+\frac{a-6}{3(a+3)}\cdot\frac{135}{a(6-a)}$$
$$\frac{5a}{a+3}+\frac{(a-6)\cdot 135}{3(a+3)\cdot a(6-a)}=\frac{5a}{a+3}-\frac{45}{a(a+3)}$$
$$\frac{5a^2-45}{a(a+3)}=\frac{5(a^2-9)}{a(a+3)}=\frac{5(a-3)(a+3)}{a(a+3)}=\frac{5(a-3)}{a}$$
$$\left(\frac{3m}{m+5}-\frac{8m}{m^2+10m+25}\right):\frac{3m+7}{m^2-25}+\frac{5m-25}{m+5}$$
$$\frac{3m}{m+5}-\frac{8m}{(m+5)^2}=\frac{3m(m+5)-8m}{(m+5)^2}=\frac{m(3m+7)}{(m+5)^2}$$
$$\frac{m(3m+7)}{(m+5)^2}:\frac{3m+7}{(m-5)(m+5)}+\frac{5(m-5)}{m+5}$$
$$\frac{m(3m+7)}{(m+5)^2}\cdot\frac{(m-5)(m+5)}{3m+7}+\frac{5(m-5)}{m+5}$$
$$\frac{m(m-5)}{m+5}+\frac{5(m-5)}{m+5}=\frac{(m-5)(m+5)}{m+5}=m-5$$
$$\left(\frac{y^2}{x^3-xy^2}+\frac{1}{x+y}\right):\left(\frac{x-y}{x^2+xy}-\frac{x}{xy+y^2}\right)$$
$$\frac{y^2}{x(x-y)(x+y)}+\frac{1}{x+y}=\frac{y^2+x(x-y)}{x(x-y)(x+y)}=\frac{x^2-xy+y^2}{x(x-y)(x+y)}$$
$$\frac{x-y}{x(x+y)}-\frac{x}{y(x+y)}=\frac{y(x-y)-x^2}{xy(x+y)}=\frac{xy-y^2-x^2}{xy(x+y)}=-\frac{x^2-xy+y^2}{xy(x+y)}$$
$$\frac{x^2-xy+y^2}{x(x-y)(x+y)}:\left(-\frac{x^2-xy+y^2}{xy(x+y)}\right)=-\frac{y}{x-y}=\frac{y}{y-x}$$
$$\left(\frac{a}{a-4}-\frac{a}{a+4}-\frac{a^2+16}{16-a^2}\right):\frac{4a+a^2}{(4-a)^2}$$
$$\frac{a}{a-4}-\frac{a}{a+4}+\frac{a^2+16}{a^2-16}$$
$$\frac{a(a+4)-a(a-4)+a^2+16}{a^2-16}=\frac{a^2+8a+16}{a^2-16}=\frac{(a+4)^2}{(a-4)(a+4)}=\frac{a+4}{a-4}$$
$$\frac{a+4}{a-4}:\frac{a(a+4)}{(4-a)^2}=\frac{a+4}{a-4}\cdot\frac{(4-a)^2}{a(a+4)}=\frac{a-4}{a}$$