Упр.542 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) b^2/(a^2-2ab):(2ab/(a^2-4b^2)-b/(a+2b));
2) (xy/(x^2-y^2)-y/(2x-2y)):3y/(x^2-y^2);
3) (2xy/(x^2-9y^2)-y/(x-3y)):y^2/(x^2+3xy);
4) ((2a+1)/(2a-1)-(2a-1)/(2a+1))·(10a-5)/(4a).

1. $$\frac{b^2}{a^2-2ab}:\left(\frac{2ab}{a^2-4b^2}-\frac{b}{a+2b}\right)$$
   $$=\frac{b^2}{a(a-2b)}:\left(\frac{2ab-b(a-2b)}{(a-2b)(a+2b)}\right)$$
   $$=\frac{b^2}{a(a-2b)}:\frac{b(a+2b)}{(a-2b)(a+2b)}$$
   $$=\frac{b^2}{a(a-2b)}\cdot\frac{(a-2b)(a+2b)}{b(a+2b)}=\frac{b}{a}.$$

2. $$\left(\frac{xy}{x^2-y^2}-\frac{y}{2x-2y}\right):\frac{3y}{x^2-y^2}$$
   $$=\left(\frac{xy}{(x-y)(x+y)}-\frac{y}{2(x-y)}\right):\frac{3y}{(x-y)(x+y)}$$
   $$=\left(\frac{2xy-y(x+y)}{2(x-y)(x+y)}\right)\cdot\frac{(x-y)(x+y)}{3y}$$
   $$=\frac{2xy-xy-y^2}{2\cdot 3y}=\frac{xy-y^2}{6y}=\frac{y(x-y)}{6y}=\frac{x-y}{6}.$$

3. $$\left(\frac{2xy}{x^2-9y^2}-\frac{y}{x-3y}\right):\frac{y^2}{x^2+3xy}$$
   $$=\left(\frac{2xy}{(x-3y)(x+3y)}-\frac{y(x+3y)}{(x-3y)(x+3y)}\right):\frac{y^2}{x(x+3y)}$$
   $$=\frac{2xy-y(x+3y)}{(x-3y)(x+3y)}\cdot\frac{x(x+3y)}{y^2}$$
   $$=\frac{2xy-xy-3y^2}{x-3y}\cdot\frac{x}{y^2}$$
   $$=\frac{xy-3y^2}{x-3y}\cdot\frac{x}{y^2}=\frac{y(x-3y)}{x-3y}\cdot\frac{x}{y^2}=\frac{x}{y}.$$

4. $$\left(\frac{2a+1}{2a-1}-\frac{2a-1}{2a+1}\right)\cdot\frac{10a-5}{4a}$$
   $$=\frac{(2a+1)^2-(2a-1)^2}{(2a-1)(2a+1)}\cdot\frac{5(2a-1)}{4a}$$
   $$=\frac{(4a^2+4a+1)-(4a^2-4a+1)}{(2a-1)(2a+1)}\cdot\frac{5(2a-1)}{4a}$$
   $$=\frac{8a}{(2a-1)(2a+1)}\cdot\frac{5(2a-1)}{4a}=\frac{10}{2a+1}.$$

Ответ

1) $$\frac{b}{a}$$; 2) $$\frac{x-y}{6}$$; 3) $$\frac{x}{y}$$; 4) $$\frac{10}{2a+1}$$.