Упр.178 ГДЗ Мерзляк Полонский 8 класс (Алгебра)

Задача

1) (15/(x-7) — x — 7) * (7-x)/(x2-16x+64);
2) (a — (5a-16)/(a-3)) : (2a — 2a/(a-3)); Выполните действия:
1) (a+2)/(a2-2a+1) : (a2-4)/(3a-3) — 3/(a-2);
2) (b2+3b)/(b3+9b) * ((b-3)/(b+3) + (b+3)/(b-3));

Подробный ответ

$$\left(\frac{15}{x-7}-x-7\right)\cdot \frac{7-x}{x^2-16x+64}$$
$$x^2-16x+64=(x-8)^2,\qquad 7-x=-(x-7)$$
$$\left(\frac{15-(x-7)(x+7)}{x-7}\right)\cdot \frac{-(x-7)}{(x-8)^2}$$
$$=\frac{15-(x^2-49)}{x-7}\cdot \frac{-(x-7)}{(x-8)^2}$$
$$=\frac{64-x^2}{x-7}\cdot \frac{-(x-7)}{(x-8)^2}$$
$$=\frac{-(x-8)(x+8)}{x-7}\cdot \frac{-(x-7)}{(x-8)^2}=\frac{x+8}{x-8}$$
$$\left(a-\frac{5a-16}{a-3}\right):\left(2a-\frac{2a}{a-3}\right)$$
$$a\ne 3$$
$$a-\frac{5a-16}{a-3}=\frac{a(a-3)-5a+16}{a-3}=\frac{a^2-8a+16}{a-3}=\frac{(a-4)^2}{a-3}$$
$$2a-\frac{2a}{a-3}=\frac{2a(a-3)-2a}{a-3}=\frac{2a^2-8a}{a-3}=\frac{2a(a-4)}{a-3}$$
$$\frac{(a-4)^2}{a-3}:\frac{2a(a-4)}{a-3}=\frac{a-4}{2a}$$
$$\frac{a+2}{a^2-2a+1}:\frac{a^2-4}{3a-3}-\frac{3}{a-2}$$
$$a^2-2a+1=(a-1)^2,\qquad a^2-4=(a-2)(a+2),\qquad 3a-3=3(a-1)$$
$$\frac{a+2}{(a-1)^2}\cdot \frac{3(a-1)}{(a-2)(a+2)}-\frac{3}{a-2}$$
$$=\frac{3}{(a-1)(a-2)}-\frac{3}{a-2}$$
$$=\frac{3-3(a-1)}{(a-1)(a-2)}=\frac{6-3a}{(a-1)(a-2)}$$
$$=-\frac{3(a-2)}{(a-1)(a-2)}=\frac{3}{1-a}$$
$$\frac{b^2+3b}{b^3+9b}\cdot \left(\frac{b-3}{b+3}+\frac{b+3}{b-3}\right)$$
$$b^3+9b=b(b^2+9)$$
$$\frac{b(b+3)}{b(b^2+9)}\cdot \frac{(b-3)^2+(b+3)^2}{(b-3)(b+3)}$$
$$=\frac{b+3}{b^2+9}\cdot \frac{2b^2+18}{(b-3)(b+3)}$$
$$=\frac{b+3}{b^2+9}\cdot \frac{2(b^2+9)}{(b-3)(b+3)}=\frac{2}{b-3}$$
$$\left(\frac{3c+1}{3c-1}-\frac{3c-1}{3c+1}\right):\frac{2c}{6c+2}$$
$$6c+2=2(3c+1)$$
$$\frac{(3c+1)^2-(3c-1)^2}{(3c-1)(3c+1)}:\frac{2c}{2(3c+1)}$$
$$=\frac{12c}{(3c-1)(3c+1)}\cdot \frac{3c+1}{c}=\frac{12}{3c-1}$$
$$\left(\frac{1}{a^2-4ab+4b^2}-\frac{1}{4b^2-a^2}\right):\frac{2a}{a^2-4b^2}$$
$$a^2-4ab+4b^2=(a-2b)^2,\qquad 4b^2-a^2=-(a-2b)(a+2b),\qquad a^2-4b^2=(a-2b)(a+2b)$$
$$\left(\frac{1}{(a-2b)^2}+\frac{1}{(a-2b)(a+2b)}\right)\cdot \frac{(a-2b)(a+2b)}{2a}$$
$$=\frac{a+2b+a-2b}{(a-2b)^2(a+2b)}\cdot \frac{(a-2b)(a+2b)}{2a}$$
$$=\frac{2a}{(a-2b)^2(a+2b)}\cdot \frac{(a-2b)(a+2b)}{2a}=\frac{1}{a-2b}$$
$$\left(\frac{a-8}{a^2-10a+25}-\frac{a}{a^2-25}\right):\frac{a-20}{(a-5)^2}$$
$$a^2-10a+25=(a-5)^2,\qquad a^2-25=(a-5)(a+5)$$
$$\left(\frac{a-8}{(a-5)^2}-\frac{a}{(a-5)(a+5)}\right)\cdot \frac{(a-5)^2}{a-20}$$
$$=\frac{(a-8)(a+5)-a(a-5)}{(a-5)^2(a+5)}\cdot \frac{(a-5)^2}{a-20}$$
$$=\frac{a^2-3a-40-a^2+5a}{(a+5)(a-20)}=\frac{2a-40}{(a+5)(a-20)}$$
$$=\frac{2(a-20)}{(a+5)(a-20)}=\frac{2}{a+5}$$
$$\left(\frac{2x+1}{x^2+6x+9}-\frac{x-2}{x^2+3x}\right):\frac{x^2+6}{x^3-9x}$$
$$x^2+6x+9=(x+3)^2,\qquad x^2+3x=x(x+3),\qquad x^3-9x=x(x-3)(x+3)$$
$$\left(\frac{2x+1}{(x+3)^2}-\frac{x-2}{x(x+3)}\right)\cdot \frac{x(x-3)(x+3)}{x^2+6}$$
$$=\frac{x(2x+1)-(x-2)(x+3)}{x(x+3)^2}\cdot \frac{x(x-3)(x+3)}{x^2+6}$$
$$=\frac{2x^2+x-(x^2+x-6)}{x(x+3)^2}\cdot \frac{x(x-3)(x+3)}{x^2+6}$$
$$=\frac{x^2+6}{x(x+3)^2}\cdot \frac{x(x-3)(x+3)}{x^2+6}=\frac{x-3}{x+3}$$