Контрольная работа 2 Вариант 1 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)

1.

1. $$\frac{56x^3y^4}{z^5}\cdot\left(-\frac{z^4}{16x^2y^6}\right) =-\frac{56}{16}\cdot\frac{x^3}{x^2}\cdot\frac{y^4}{y^6}\cdot\frac{z^4}{z^5} =-\frac{7x}{2y^2z}.$$
2. $$\frac{72a^7}{c^{10}}:\left(24a^3c^8\right) =\frac{72a^7}{c^{10}\cdot24a^3c^8} =\frac{3a^4}{c^{18}}.$$
3. $$\frac{3b-3c}{c}\cdot\frac{4c^2}{b^2-c^2} =\frac{3(b-c)}{c}\cdot\frac{4c^2}{(b-c)(b+c)} =\frac{12c}{b+c}.$$
4. $$\frac{6x-30}{x+8}:\frac{x^2-25}{2x+16} =\frac{6(x-5)}{x+8}\cdot\frac{2(x+8)}{(x-5)(x+5)} =\frac{12}{x+5}.$$

2.

1. $$\frac{2a}{a-2}+\frac{a+7}{8-4a}\cdot\frac{32}{7a+a^2} =\frac{2a}{a-2}+\frac{a+7}{4(2-a)}\cdot\frac{32}{a(a+7)}$$
   $$=\frac{2a}{a-2}-\frac{8}{a(a-2)} =\frac{2a^2-8}{a(a-2)} =\frac{2(a^2-4)}{a(a-2)} =\frac{2(a-2)(a+2)}{a(a-2)} =\frac{2(a+2)}{a}.$$
2. $$\left(\frac{a-1}{a+1}-\frac{a+1}{a-1}\right):\frac{2a}{1-a^2}$$
   $$=\frac{(a-1)^2-(a+1)^2}{(a-1)(a+1)}\cdot\frac{1-a^2}{2a} =\frac{-4a}{a^2-1}\cdot\frac{1-a^2}{2a} =2.$$

3.

$$\left(\frac{b^3}{b^2-8b+16}-\frac{b^2}{b-4}\right):\left(\frac{b^2}{b^2-16}-\frac{b}{b-4}\right)$$
$$=\left(\frac{b^3}{(b-4)^2}-\frac{b^2}{b-4}\right):\left(\frac{b^2}{(b-4)(b+4)}-\frac{b}{b-4}\right)$$
$$=\frac{b^3-b^2(b-4)}{(b-4)^2}:\frac{b^2-b(b+4)}{(b-4)(b+4)}$$
$$=\frac{4b^2}{(b-4)^2}:\frac{-4b}{(b-4)(b+4)}$$
$$=\frac{4b^2}{(b-4)^2}\cdot\frac{(b-4)(b+4)}{-4b} =\frac{b(b+4)}{-(b-4)} =\frac{b^2+4b}{4-b}.$$

4.

Из равенства $$64x^2+\frac{1}{x^2}=65$$ получаем:
$$64x^2+16+ \frac{1}{x^2}=81$$
$$64x^2+16+\frac{1}{x^2}=65+16$$
$$\left(8x+\frac{1}{x}\right)^2=81.$$
Тогда
$$8x+\frac{1}{x}=\pm 9.$$

Ответ: 1) $$-\frac{7x}{2y^2z}$$; $$\frac{3a^4}{c^{18}}$$; $$\frac{12c}{b+c}$$; $$\frac{12}{x+5}$$; 2) $$\frac{2(a+2)}{a}$$; $$2$$; 3) верно; 4) $$\pm 9$$.