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

1) $$\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}\right):\frac{12a^2}{4-a^2}$$

$$\frac{(a-2)^2-(a+2)^2}{(a+2)(a-2)}:\frac{12a^2}{4-a^2}$$

$$\frac{a^2-4a+4-a^2-4a-4}{a^2-4}:\frac{12a^2}{4-a^2}$$

$$\frac{-8a}{a^2-4}\cdot\frac{4-a^2}{12a^2}=\frac{-8a\cdot\bigl(-(a^2-4)\bigr)}{(a^2-4)\cdot 12a^2}=\frac{2}{3a}$$

2) $$\left(\frac{8x}{x-2}+2x\right):\frac{4x+8}{7x-14}$$

$$\frac{8x+2x(x-2)}{x-2}:\frac{4x+8}{7x-14}$$

$$\frac{2x^2+4x}{x-2}\cdot\frac{7(x-2)}{4(x+2)}=\frac{2x(x+2)\cdot 7(x-2)}{(x-2)\cdot 4(x+2)}=\frac{7x}{2}$$

3) $$\frac{5a}{a+3}+\frac{a-6}{3a+9}\cdot\frac{135}{6a-a^2}$$

$$\frac{5a}{a+3}+\frac{(a-6)\cdot 135}{3(a+3)\cdot a(6-a)}$$

$$\frac{5a}{a+3}+\frac{-45}{a(a+3)}=\frac{5a^2-45}{a(a+3)}=\frac{5(a^2-9)}{a(a+3)}=\frac{5(a-3)(a+3)}{a(a+3)}=\frac{5(a-3)}{a}$$

4) $$\left(\frac{3m}{m+5}-\frac{8m}{m^2+10m+25}\right):\frac{3m+7}{m^2-25}+\frac{5m-25}{m+5}$$

$$\left(\frac{3m}{m+5}-\frac{8m}{(m+5)^2}\right)\cdot\frac{m^2-25}{3m+7}+\frac{5m-25}{m+5}$$

$$\frac{3m(m+5)-8m}{(m+5)^2}\cdot\frac{(m-5)(m+5)}{3m+7}+\frac{5m-25}{m+5}$$

$$\frac{3m^2+7m}{(m+5)^2}\cdot\frac{(m-5)(m+5)}{3m+7}+\frac{5m-25}{m+5}$$

$$\frac{m(3m+7)(m-5)}{(m+5)(3m+7)}+\frac{5m-25}{m+5}=\frac{m(m-5)}{m+5}+\frac{5m-25}{m+5}$$

$$\frac{m^2-5m+5m-25}{m+5}=\frac{m^2-25}{m+5}=\frac{(m-5)(m+5)}{m+5}=m-5$$

5) $$\left(\frac{y^2}{x^3-xy^2}+\frac{1}{x+y}\right):\left(\frac{x-y}{x^2+xy}-\frac{x}{xy+y^2}\right)$$

$$\left(\frac{y^2}{x(x-y)(x+y)}+\frac{1}{x+y}\right):\left(\frac{x-y}{x(x+y)}-\frac{x}{y(x+y)}\right)$$

$$\frac{y^2+x(x-y)}{x(x-y)(x+y)}:\frac{y(x-y)-x^2}{xy(x+y)}$$

$$\frac{y^2+x^2-xy}{x(x^2-y^2)}:\frac{xy-y^2-x^2}{xy(x+y)}$$

$$\frac{y^2-xy+x^2}{x(x-y)(x+y)}:\frac{-(y^2-xy+x^2)}{xy(x+y)}$$

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

6) $$\left(\frac{a}{a-4}-\frac{a}{a+4}-\frac{a^2+16}{16-a^2}\right):\frac{4a+a^2}{(4-a)^2}$$

$$\left(\frac{a}{a-4}-\frac{a}{a+4}+\frac{a^2+16}{a^2-16}\right)\cdot\frac{(4-a)^2}{4a+a^2}$$

$$\frac{a(a+4)-a(a-4)+a^2+16}{a^2-16}\cdot\frac{(4-a)^2}{a(a+4)}$$

$$\frac{a^2+4a-a^2+4a+a^2+16}{a^2-16}\cdot\frac{(4-a)^2}{a(a+4)}=\frac{a^2+8a+16}{a^2-16}\cdot\frac{(4-a)^2}{a(a+4)}$$

$$\frac{(a+4)^2}{(a-4)(a+4)}\cdot\frac{(4-a)^2}{a(a+4)}=\frac{a-4}{a}$$

Ответ: 1) $$\frac{2}{3a}$$; 2) $$\frac{7x}{2}$$; 3) $$\frac{5(a-3)}{a}$$; 4) $$m-5$$; 5) $$\frac{y}{y-x}$$; 6) $$\frac{a-4}{a}$$.