Упр.948 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б) a2/(a2+4a+4) * (a2-4)/(a3-2a2) + (a5-8a2)/a : (a2-4);
в) ((m+2)/(8-8m+2m2)+ 1/(4-2m) — 2/(m2-4m+4))*3m — 3m;
г) (2/(a2-4a+4) — 1/(4-2a) — (a+2)/(2(2-a)2))*5a-5a;
д) (1/(2-4c) + (1+c)/(8c3-1) : (1+2c)/(4c2+2c+1)) * (4c-2)/(2c+1) — 1/(1+2c)2;
е) (1/(2-4b) + (b+1)/(8b3-1) * (4b2+2b+1)/(1+2b)): (2b+1)/(4b-2) — 1/(2b+1)2.

Подробный ответ

а)
$$\frac{x^5}{x^2-6x+9}\cdot\frac{x^2-9}{x^3+3x^2}-\frac{3x^5+81x^2}{x^2}:(x^2-9)$$
$$=\frac{x^5}{(x-3)^2}\cdot\frac{(x-3)(x+3)}{x^2(x+3)}-\frac{3x^2(x^3+27)}{x^2}\cdot\frac{1}{x^2-9}$$
$$=\frac{x^3}{x-3}-\frac{3(x^3+27)}{(x-3)(x+3)}$$
$$=\frac{x^3-3(x^2-3x+9)}{x-3}$$
$$=\frac{x^3-3x^2+9x-27}{x-3}$$
$$=\frac{(x-3)(x^2+9)}{x-3}=x^2+9.$$
б)
$$\frac{a^2}{a^2+4a+4}\cdot\frac{a^2-4}{a^3-2a^2}+\frac{a^5-8a^2}{a}:(a^2-4)$$
$$=\frac{a^2}{(a+2)^2}\cdot\frac{(a-2)(a+2)}{a^2(a-2)}+\frac{a^2(a^3-8)}{a}\cdot\frac{1}{a^2-4}$$
$$=\frac{1}{a+2}+\frac{a(a^2+2a+4)}{a+2}$$
$$=\frac{1+a^3+2a^2+4a}{a+2}.$$
в)
$$\left(\frac{m+2}{8-8m+2m^2}+\frac{1}{4-2m}-\frac{2}{m^2-4m+4}\right)\cdot 3m-3m$$
$$=\left(\frac{m+2}{2(2-m)^2}+\frac{1}{2(2-m)}-\frac{2}{(2-m)^2}\right)\cdot 3m-3m$$
$$=\left(\frac{m+2+2(2-m)-4}{2(2-m)^2}\right)\cdot 3m-3m$$
$$=0\cdot 3m-3m=-3m.$$
г)
$$\left(\frac{2}{a^2-4a+4}-\frac{1}{4-2a}-\frac{a+2}{2(2-a)^2}\right)\cdot 5a-5a$$
$$=\left(\frac{2}{(2-a)^2}-\frac{1}{2(2-a)}-\frac{a+2}{2(2-a)^2}\right)\cdot 5a-5a$$
$$=\left(\frac{4-(2-a)-(a+2)}{2(2-a)^2}\right)\cdot 5a-5a$$
$$=0\cdot 5a-5a=-5a.$$
д)
$$\left(\frac{1}{2-4c}+\frac{1+c}{8c^3-1}:\frac{1+2c}{4c^2+2c+1}\right)\cdot\frac{4c-2}{2c+1}-\frac{1}{(1+2c)^2}$$
$$=\left(\frac{1}{2(1-2c)}+\frac{1+c}{(2c-1)(4c^2+2c+1)}\cdot\frac{4c^2+2c+1}{1+2c}\right)\cdot\frac{4c-2}{2c+1}-\frac{1}{(1+2c)^2}$$
$$=\left(\frac{1}{2(1-2c)}-\frac{1+c}{(1-2c)(1+2c)}\right)\cdot\frac{2(2c-1)}{2c+1}-\frac{1}{(1+2c)^2}$$
$$=\frac{1+2c-2(1+c)}{2(1-2c)(1+2c)}\cdot\frac{-2(1-2c)}{2c+1}-\frac{1}{(1+2c)^2}$$
$$=\frac{1}{(1+2c)^2}-\frac{1}{(1+2c)^2}=0.$$
е)
$$\left(\frac{1}{2-4b}+\frac{b+1}{8b^3-1}\cdot\frac{4b^2+2b+1}{1+2b}\right):\frac{2b+1}{4b-2}-\frac{1}{(2b+1)^2}$$
$$=\left(\frac{1}{2(1-2b)}+\frac{b+1}{(2b-1)(4b^2+2b+1)}\cdot\frac{4b^2+2b+1}{1+2b}\right)\cdot\frac{4b-2}{2b+1}-\frac{1}{(2b+1)^2}$$
$$=\left(\frac{1}{2(1-2b)}-\frac{b+1}{(1-2b)(1+2b)}\right)\cdot\frac{2(2b-1)}{2b+1}-\frac{1}{(2b+1)^2}$$
$$=\frac{1+2b-2(b+1)}{2(1-2b)(1+2b)}\cdot\frac{-2(1-2b)}{2b+1}-\frac{1}{(2b+1)^2}$$
$$=\frac{1}{(2b+1)^2}-\frac{1}{(2b+1)^2}=0.$$