Упр.151 Повторение ГДЗ Мерзляк 11 класс Базовый уровень (Алгебра)

1) (4n+5m)/m-(6n^2+5m^2)/(mn);
2) (a+2)/(3a-3)+(3-a)/(5a-5);
3) (x-5)/(x+5)-(x-1)/(x-5);
4) 4b/(3b-24)+3b/(16-2b);
5) 3p/(3p+q)-9p^2/(9p^2+6pq+q^2);
6) 4/(m^2-36)-2/(m^2-6m);
7) 8-(3a+8c)/c;
8) (m^2-n^2)/(m+4n)+m-4n;
9) x-49/(x-7)-7.

1. $$\frac{4n+5m}{m}-\frac{6n^2+5m^2}{mn}=\frac{n(4n+5m)-6n^2-5m^2}{mn}$$
   $$=\frac{4n^2+5mn-6n^2-5m^2}{mn}=\frac{-5m^2+5mn-2n^2}{mn}.$$
2. $$\frac{a+2}{3a-3}+\frac{3-a}{5a-5}=\frac{a+2}{3(a-1)}+\frac{3-a}{5(a-1)}$$
   $$=\frac{5(a+2)+3(3-a)}{15(a-1)}=\frac{5a+10+9-3a}{15(a-1)}$$
   $$=\frac{2a+19}{15(a-1)}=\frac{2a+19}{15a-15}.$$
3. $$\frac{x-5}{x+5}-\frac{x-1}{x-5}=\frac{(x-5)^2-(x-1)(x+5)}{(x+5)(x-5)}$$
   $$=\frac{x^2-10x+25-x^2-4x+5}{x^2-25}=\frac{30-14x}{x^2-25}.$$
4. $$\frac{4b}{3b-24}+\frac{3b}{16-2b}=\frac{4b}{3(b-8)}+\frac{3b}{-2(b-8)}$$
   $$=\frac{8b-9b}{6(b-8)}=\frac{-b}{6(b-8)}=\frac{b}{6(8-b)}=\frac{b}{48-6b}.$$
5. $$\frac{3p}{3p+q}-\frac{9p^2}{9p^2+6pq+q^2}=\frac{3p}{3p+q}-\frac{9p^2}{(3p+q)^2}$$
   $$=\frac{3p(3p+q)-9p^2}{(3p+q)^2}=\frac{9p^2+3pq-9p^2}{(3p+q)^2}$$
   $$=\frac{3pq}{(3p+q)^2}.$$
6. $$\frac{4}{m^2-36}-\frac{2}{m^2-6m}=\frac{4}{(m-6)(m+6)}-\frac{2}{m(m-6)}$$
   $$=\frac{4m-2(m+6)}{m(m-6)(m+6)}=\frac{2(m-6)}{m(m-6)(m+6)}$$
   $$=\frac{2}{m(m+6)}.$$
7. $$8-\frac{3a+8c}{c}=\frac{8c-(3a+8c)}{c}=-\frac{3a}{c}.$$
8. $$\frac{m^2-n^2}{m+4n}+m-4n=\frac{m^2-n^2+(m-4n)(m+4n)}{m+4n}$$
   $$=\frac{m^2-n^2+m^2-16n^2}{m+4n}=\frac{2m^2-17n^2}{m+4n}.$$
9. $$x-\frac{49}{x-7}-7=\frac{x(x-7)-49-7(x-7)}{x-7}$$
   $$=\frac{x^2-7x-49-7x+49}{x-7}=\frac{x^2-14x}{x-7}.$$

Ответ

1. $$\frac{-5m^2+5mn-2n^2}{mn}$$
2. $$\frac{2a+19}{15a-15}$$
3. $$\frac{30-14x}{x^2-25}$$
4. $$\frac{b}{48-6b}$$
5. $$\frac{3pq}{(3p+q)^2}$$
6. $$\frac{2}{m(m+6)}$$
7. $$-\frac{3a}{c}$$
8. $$\frac{2m^2-17n^2}{m+4n}$$
9. $$\frac{x^2-14x}{x-7}$$