Упр.510 ГДЗ Колягин Ткачёва 7 класс (Алгебра)

1) 16ab^2-5b^2 c-10c^3+32ac^2;
2) 6mnk^2+15m^2 k-14n^3 k-35mn^2;
3) -28ac+35c^2-10cx+8ax;
4) -24bx-15c^2+40bc+9cx.
Выполнить действия:
1) (a+b)/a-a/(a-b)-b^2/(a^2-ab);
2) (5b-1)/(3b^2-3)+(b+2)/(2b+2)-(b+1)/(b-1);
3) 6a/(9a^2-1)+(3a+1)/(3-9a)+(3a-1)/(6a+2);
4) 7/m-4/(m-2n)-(m-n)/(4n^2-m^2 );
5) x-xy/(x+y)-x^3/(x^2-y^2 );
6) a-2+4a/(2+a)-(a^3+b)/(a^2+2a).

1. $$16ab^2-5b^2c-10c^3+32ac^2=(16ab^2+32ac^2)-(5b^2c+10c^3)$$

   $$=16a(b^2+2c^2)-5c(b^2+2c^2)=(b^2+2c^2)(16a-5c).$$

2. $$6mnk^2+15m^2k-14n^3k-35mn^2=(6mnk^2+15m^2k)-(14n^3k+35mn^2)$$

   $$=3mk(2nk+5m)-7n^2(2nk+5m)=(2nk+5m)(3mk-7n^2).$$

3. $$-28ac+35c^2-10cx+8ax=(8ax-28ac)+(35c^2-10cx)$$

   $$=4a(2x-7c)+5c(7c-2x)=4a(2x-7c)-5c(2x-7c)$$

   $$=(2x-7c)(4a-5c).$$

4. $$-24bx-15c^2+40bc+9cx=(40bc-24bx)-(15c^2-9cx)$$

   $$=8b(5c-3x)-3c(5c-3x)=(5c-3x)(8b-3c).$$

1. $$\frac{a+b}{a}-\frac{a}{a-b}-\frac{b^2}{a^2-ab}$$

   ОДЗ: $$a\ne 0,\ a\ne b.$$

   $$\frac{a+b}{a}-\frac{a}{a-b}-\frac{b^2}{a(a-b)}$$

   $$=\frac{(a+b)(a-b)-a^2-b^2}{a(a-b)}$$

   $$=\frac{a^2-b^2-a^2-b^2}{a(a-b)}=\frac{-2b^2}{a(a-b)}=\frac{2b^2}{a(b-a)}.$$

2. $$\frac{5b-1}{3b^2-3}+\frac{b+2}{2b+2}-\frac{b+1}{b-1}$$

   ОДЗ: $$b\ne 1,\ b\ne -1.$$

   $$\frac{5b-1}{3(b-1)(b+1)}+\frac{b+2}{2(b+1)}-\frac{b+1}{b-1}$$

   $$=\frac{2(5b-1)+3(b+2)(b-1)-6(b+1)^2}{6(b-1)(b+1)}$$

   $$=\frac{10b-2+3(b^2+b-2)-6(b^2+2b+1)}{6(b^2-1)}$$

   $$=\frac{-3b^2+b-14}{6(b^2-1)}.$$

3. $$\frac{6a}{9a^2-1}+\frac{3a+1}{3-9a}+\frac{3a-1}{6a+2}$$

   ОДЗ: $$a\ne \frac13,\ a\ne -\frac13.$$

   $$\frac{6a}{(3a-1)(3a+1)}-\frac{3a+1}{3(3a-1)}+\frac{3a-1}{2(3a+1)}$$

   $$=\frac{36a-2(3a+1)^2+3(3a-1)^2}{6(3a-1)(3a+1)}$$

   $$=\frac{9a^2+6a+1}{6(9a^2-1)}=\frac{(3a+1)^2}{6(3a-1)(3a+1)}=\frac{3a+1}{6(3a-1)}.$$

4. $$\frac{7}{m}-\frac{4}{m-2n}-\frac{m-n}{4n^2-m^2}$$

   ОДЗ: $$m\ne 0,\ m\ne 2n,\ m\ne -2n.$$

   $$\frac{7}{m}-\frac{4}{m-2n}+\frac{m-n}{(m-2n)(m+2n)}$$

   $$=\frac{7(m^2-4n^2)-4m(m+2n)+m(m-n)}{m(m^2-4n^2)}$$

   $$=\frac{4m^2-28n^2-9mn}{m(m^2-4n^2)}.$$

5. $$x-\frac{xy}{x+y}-\frac{x^3}{x^2-y^2}$$

   ОДЗ: $$x\ne -y,\ x\ne y.$$

   $$x-\frac{xy}{x+y}-\frac{x^3}{(x-y)(x+y)}$$

   $$=\frac{x(x^2-y^2)-xy(x-y)-x^3}{x^2-y^2}$$

   $$=\frac{-x^2y}{x^2-y^2}=\frac{x^2y}{y^2-x^2}.$$

6. $$a-2+\frac{4a}{a+2}-\frac{a^3+b}{a^2+2a}$$

   ОДЗ: $$a\ne 0,\ a\ne -2.$$

   $$a-2+\frac{4a}{a+2}-\frac{a^3+b}{a(a+2)}$$

   $$=\frac{a(a-2)(a+2)+4a^2-a^3-b}{a(a+2)}$$

   $$=\frac{a^3-4a+4a^2-a^3-b}{a(a+2)}=\frac{4a^2-4a-b}{a(a+2)}.$$

Ответ

1) $$(b^2+2c^2)(16a-5c);$$
2) $$(2nk+5m)(3mk-7n^2);$$
3) $$(2x-7c)(4a-5c);$$
4) $$(5c-3x)(8b-3c).$$

1) $$\frac{2b^2}{a(b-a)};$$
2) $$\frac{-3b^2+b-14}{6(b^2-1)};$$
3) $$\frac{3a+1}{6(3a-1)};$$
4) $$\frac{4m^2-28n^2-9mn}{m(m^2-4n^2)};$$
5) $$\frac{x^2y}{y^2-x^2};$$
6) $$\frac{4a^2-4a-b}{a(a+2)}.$$