Упр.37.11 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) (a^2+1)/(a^2-2a+1)+(a+1)/(a-1);
2) (a^2+b^2)/(a^2-b^2 )-(a-b)/(a+b);
3) (c+7)/(c-7)+28c/(49-c^2 );
4) (5a+3)/(2a^2+6a)+(6-3a)/(a^2-9);
5) a/(a^2-4a+4)-(a+4)/(a^2-4);
6) 2p/(p-5)-5/(p+5)+(2p^2)/(25-p^2 );
7) 1/y-(y+8)/(16-y^2 )-2/(y-4);
8) (2b-1)/(4b+2)+4b/(4b^2-1)+(2b+1)/(3-6b).

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

Ответ

1) $$\frac{2a^2}{(a-1)^2}$$; 2) $$\frac{2ab}{a^2-b^2}$$; 3) $$\frac{c-7}{c+7}$$; 4) $$-\frac{a^2+9}{2a(a^2-9)}$$; 5) $$\frac{8}{(a-2)^2(a+2)}$$; 6) $$\frac{5}{p-5}$$; 7) $$\frac{16}{y(16-y^2)}$$; 8) $$\frac{2b+1}{6(2b-1)}$$.