Упр.110 ГДЗ Мерзляк Полонский 8 класс (Алгебра)

Задача

1) (a2+1)/(a2-2a+1) + (a+1)(a-1);
2) (a2+b2)/(a2-b2) — (a-b)/(a+b); Представьте в виде дроби выражение:
1) a/b + 1;
2) x/y — x;
3) m/n + n/m + 2;
4) 9/p2 — 4/p + 3;

Подробный ответ

$$\frac{a^2+1}{a^2-2a+1}+\frac{a+1}{a-1}=\frac{a^2+1}{(a-1)^2}+\frac{a+1}{a-1}$$
$$=\frac{a^2+1+(a+1)(a-1)}{(a-1)^2}=\frac{a^2+1+a^2-1}{(a-1)^2}=\frac{2a^2}{(a-1)^2}$$
$$\frac{a^2+b^2}{a^2-b^2}-\frac{a-b}{a+b}=\frac{a^2+b^2}{(a-b)(a+b)}-\frac{(a-b)^2}{(a-b)(a+b)}$$
$$=\frac{a^2+b^2-(a-b)^2}{a^2-b^2}=\frac{a^2+b^2-a^2+2ab-b^2}{a^2-b^2}=\frac{2ab}{a^2-b^2}$$
$$\frac{c+7}{c-7}+\frac{28c}{49-c^2}=\frac{c+7}{c-7}-\frac{28c}{(c-7)(c+7)}$$
$$=\frac{(c+7)^2-28c}{(c-7)(c+7)}=\frac{c^2+14c+49-28c}{c^2-49}$$
$$=\frac{c^2-14c+49}{c^2-49}=\frac{(c-7)^2}{(c-7)(c+7)}=\frac{c-7}{c+7}$$
$$\frac{5a+3}{2a^2+6a}+\frac{6-3a}{a^2-9}=\frac{5a+3}{2a(a+3)}+\frac{6-3a}{(a-3)(a+3)}$$
$$=\frac{(5a+3)(a-3)+2a(6-3a)}{2a(a-3)(a+3)}$$
$$=\frac{5a^2-15a+3a-9+12a-6a^2}{2a(a^2-9)}=\frac{-a^2-9}{2a(a^2-9)}$$
$$\frac{a}{a^2-4a+4}-\frac{a+4}{a^2-4}=\frac{a}{(a-2)^2}-\frac{a+4}{(a-2)(a+2)}$$
$$=\frac{a(a+2)-(a+4)(a-2)}{(a-2)^2(a+2)}$$
$$=\frac{a^2+2a-a^2-2a+8}{(a-2)^2(a+2)}=\frac{8}{(a-2)^2(a+2)}$$
$$\frac{2p}{p-5}-\frac{5}{p+5}+\frac{2p^2}{25-p^2}=\frac{2p}{p-5}-\frac{5}{p+5}-\frac{2p^2}{(p-5)(p+5)}$$
$$=\frac{2p(p+5)-5(p-5)-2p^2}{(p-5)(p+5)}$$
$$=\frac{2p^2+10p-5p+25-2p^2}{(p-5)(p+5)}=\frac{5(p+5)}{(p-5)(p+5)}=\frac{5}{p-5}$$
$$\frac{1}{y}-\frac{y+8}{16-y^2}-\frac{2}{y-4}=\frac{1}{y}+\frac{y+8}{(y-4)(y+4)}-\frac{2}{y-4}$$
$$=\frac{(y-4)(y+4)+y(y+8)-2y(y+4)}{y(y-4)(y+4)}$$
$$=\frac{y^2-16+y^2+8y-2y^2-8y}{y(y^2-16)}=\frac{-16}{y(y^2-16)}=\frac{16}{y(16-y^2)}$$
$$\frac{2b-1}{4b+2}+\frac{4b}{4b^2-1}+\frac{2b+1}{3-6b}=\frac{2b-1}{2(2b+1)}+\frac{4b}{(2b-1)(2b+1)}-\frac{2b+1}{3(2b-1)}$$
$$=\frac{3(2b-1)^2+4b\cdot 6-2(2b+1)^2}{6(2b+1)(2b-1)}$$
$$=\frac{12b^2-12b+3+24b-8b^2-8b-2}{6(2b+1)(2b-1)}=\frac{4b^2+4b+1}{6(2b+1)(2b-1)}$$
$$=\frac{(2b+1)^2}{6(2b+1)(2b-1)}=\frac{2b+1}{6(2b-1)}$$