Упр.958 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б) 5a/(5a+3b) + ((5a+3b)/(5a-3b) — 25a2/(25a2-9b2))*(5a-3b)/(10a+3b) =1;
в) ((2a/(a2-16) — 4/(4+a))*(a+4)/(8-a) + a2/(32-8a) = -(a+4)/8;
г) 1/x*((y2-xy)/(x+y))2*((x+y)/(x-y)2 + (x+y)/(xy-y2)) + x/(x+y)=1;
д) m/(m2-2m+1) — 1/(1-m) * m/(m+1) — 2/(m+1) = m/(m-1)2 — (m-2)/(m2-1);
е) (a/(b2+ab) — (a-b)/(a2+ab)): (b2/(a3-ab2) + 1/(a+b))=a/b-1.

Подробный ответ

а)
$$\left(\frac{m^2+2m}{4m^2-n^2}-\frac{1}{2m+n}\right):\frac{m^2+n}{20m^2+10mn}+\frac{5n}{n-2m}$$
$$=\left(\frac{m^2+2m}{(2m-n)(2m+n)}-\frac{1}{2m+n}\right):\frac{m^2+n}{10m(2m+n)}+\frac{5n}{n-2m}$$
$$=\frac{m^2+2m-2m+n}{(2m-n)(2m+n)}\cdot\frac{10m(2m+n)}{m^2+n}+\frac{5n}{n-2m}$$
$$=\frac{m^2+n}{2m-n}\cdot\frac{10m}{m^2+n}+\frac{5n}{n-2m}$$
$$=\frac{10m}{2m-n}+\frac{5n}{n-2m}=\frac{10m}{2m-n}-\frac{5n}{2m-n}$$
$$=\frac{10m-5n}{2m-n}=\frac{5(2m-n)}{2m-n}=5.$$
б)
$$\frac{5a}{5a+3b}+\left(\frac{5a+3b}{5a-3b}-\frac{25a^2}{25a^2-9b^2}\right)\cdot\frac{5a-3b}{10a+3b}$$
$$=\frac{5a}{5a+3b}+\left(\frac{5a+3b}{5a-3b}-\frac{25a^2}{(5a-3b)(5a+3b)}\right)\cdot\frac{5a-3b}{10a+3b}$$
$$=\frac{5a}{5a+3b}+\frac{(5a+3b)^2-25a^2}{(5a-3b)(5a+3b)}\cdot\frac{5a-3b}{10a+3b}$$
$$=\frac{5a}{5a+3b}+\frac{30ab+9b^2}{(5a-3b)(5a+3b)}\cdot\frac{5a-3b}{10a+3b}$$
$$=\frac{5a}{5a+3b}+\frac{3b(10a+3b)}{(5a-3b)(5a+3b)}\cdot\frac{5a-3b}{10a+3b}$$
$$=\frac{5a}{5a+3b}+\frac{3b}{5a+3b}=\frac{5a+3b}{5a+3b}=1.$$
в)
$$\left(\frac{2a}{a^2-16}-\frac{4}{4+a}\right)\cdot\frac{a+4}{8-a}+\frac{a^2}{32-8a}$$
$$=\left(\frac{2a}{(a-4)(a+4)}-\frac{4}{a+4}\right)\cdot\frac{a+4}{8-a}+\frac{a^2}{8(4-a)}$$
$$=\frac{2a-4(a-4)}{(a-4)(a+4)}\cdot\frac{a+4}{8-a}+\frac{a^2}{8(4-a)}$$
$$=\frac{16-2a}{(a-4)(a+4)}\cdot\frac{a+4}{8-a}+\frac{a^2}{8(4-a)}$$
$$=\frac{2(8-a)}{(a-4)(a+4)}\cdot\frac{a+4}{8-a}+\frac{a^2}{8(4-a)}$$
$$=\frac{2}{a-4}-\frac{a^2}{8(a-4)}=\frac{16-a^2}{8(a-4)}$$
$$=\frac{-(a-4)(a+4)}{8(a-4)}=-\frac{a+4}{8}.$$
г)
$$\frac{1}{x}\left(\frac{y^2-xy}{x+y}\right)^2\left(\frac{x+y}{(x-y)^2}+\frac{x+y}{xy-y^2}\right)+\frac{x}{x+y}$$
$$=\frac{1}{x}\cdot\frac{y^2(y-x)^2}{(x+y)^2}\left(\frac{x+y}{(x-y)^2}+\frac{x+y}{y(x-y)}\right)+\frac{x}{x+y}$$
$$=\frac{1}{x}\cdot\frac{y^2(x-y)^2}{(x+y)^2}\cdot\frac{y(x+y)+x(x-y)}{y(x-y)^2}+\frac{x}{x+y}$$
$$=\frac{1}{x}\cdot\frac{y^2(x-y)^2}{(x+y)^2}\cdot\frac{x+y}{y(x-y)^2}+\frac{x}{x+y}$$
$$=\frac{y}{x+y}+\frac{x}{x+y}=1.$$
д)
$$\frac{m}{m^2-2m+1}-\frac{1}{1-m}\cdot\frac{m}{m+1}-\frac{2}{m+1}$$
$$=\frac{m}{(m-1)^2}+\frac{1}{m-1}\cdot\frac{m}{m+1}-\frac{2}{m+1}$$
$$=\frac{m}{(m-1)^2}+\frac{m}{(m-1)(m+1)}-\frac{2}{m+1}$$
$$=\frac{m(m+1)+m(m-1)-2(m-1)^2}{(m-1)^2(m+1)}$$
$$=\frac{4m-2}{(m-1)^2(m+1)}.$$
Правая часть:
$$\frac{m}{(m-1)^2}-\frac{m-2}{m^2-1}=\frac{m}{(m-1)^2}-\frac{m-2}{(m-1)(m+1)}$$
$$=\frac{m(m+1)-(m-2)(m-1)}{(m-1)^2(m+1)}=\frac{4m-2}{(m-1)^2(m+1)}.$$
е)
$$\left(\frac{a}{b^2+ab}-\frac{a-b}{a^2+ab}\right):\left(\frac{b^2}{a^3-ab^2}+\frac{1}{a+b}\right)$$
$$=\left(\frac{a}{b(b+a)}-\frac{a-b}{a(a+b)}\right):\left(\frac{b^2}{a(a^2-b^2)}+\frac{1}{a+b}\right)$$
$$=\frac{a^2-b(a-b)}{ab(a+b)}:\frac{b^2+a(a-b)}{a(a-b)(a+b)}$$
$$=\frac{a^2-ab+b^2}{ab(a+b)}:\frac{a^2-ab+b^2}{a(a-b)(a+b)}$$
$$=\frac{a-b}{b}.$$