Упр.498 ГДЗ Колягин Ткачёва 7 класс (Алгебра)

1. $$c(a-b)+b(b-a)=c(a-b)-b(a-b)=(a-b)(c-b).$$

2. $$a(b-c)-c(c-b)=a(b-c)+c(b-c)=(b-c)(a+c).$$

3. $$(x-y)+b(y-x)=(x-y)-b(x-y)=(x-y)(1-b).$$

4. $$2b(x-y)-(y-x)=2b(x-y)+(x-y)=(x-y)(2b+1).$$

5. $$\left(\frac{6}{a-b}-\frac{5}{a+b}\right)\cdot \frac{a-b}{a+11b} =\frac{6(a+b)-5(a-b)}{(a-b)(a+b)}\cdot \frac{a-b}{a+11b}$$
   $$=\frac{6a+6b-5a+5b}{(a-b)(a+b)}\cdot \frac{a-b}{a+11b} =\frac{a+11b}{(a-b)(a+b)}\cdot \frac{a-b}{a+11b} =\frac{1}{a+b}.$$

6. $$\left(\frac{3}{c}+\frac{3}{c+d}\right)\cdot \frac{c}{18(2c+d)} =\frac{3(c+d)+3c}{c(c+d)}\cdot \frac{c}{18(2c+d)}$$
   $$=\frac{6c+3d}{c(c+d)}\cdot \frac{c}{18(2c+d)} =\frac{3(2c+d)}{(c+d)\cdot 18(2c+d)} =\frac{1}{6(c+d)}.$$

7. $$\frac{y-1}{y}:\left(\frac{y^2+1}{y^2+2y}-\frac{2}{y+2}\right) =\frac{y-1}{y}:\left(\frac{y^2+1}{y(y+2)}-\frac{2}{y+2}\right)$$
   $$=\frac{y-1}{y}:\left(\frac{y^2+1-2y}{y(y+2)}\right) =\frac{y-1}{y}:\left(\frac{(y-1)^2}{y(y+2)}\right)$$
   $$=\frac{y-1}{y}\cdot \frac{y(y+2)}{(y-1)^2} =\frac{y+2}{y-1}.$$

8. $$\frac{m-2}{m-5}:\left(\frac{m^2+24}{m^2-25}-\frac{4}{m-5}\right) =\frac{m-2}{m-5}:\left(\frac{m^2+24-4(m+5)}{m^2-25}\right)$$
   $$=\frac{m-2}{m-5}:\left(\frac{m^2-4m+4}{m^2-25}\right) =\frac{m-2}{m-5}:\left(\frac{(m-2)^2}{(m-5)(m+5)}\right)$$
   $$=\frac{m-2}{m-5}\cdot \frac{(m-5)(m+5)}{(m-2)^2} =\frac{m+5}{m-2}.$$

Ответ

1) $$(a-b)(c-b);$$
2) $$(b-c)(a+c);$$
3) $$(x-y)(1-b);$$
4) $$(x-y)(2b+1);$$
5) $$\frac{1}{a+b};$$
6) $$\frac{1}{6(c+d)};$$
7) $$\frac{y+2}{y-1};$$
8) $$\frac{m+5}{m-2}.$$