Помогите с номером 289,как его?

Решите задачу:

$(1+\frac{2}{a}):(1-\frac{2}{a})= \frac{a+2}{a}:\frac{a-2}{a}=\frac{a+2}{a}* \frac{a}{a-2}= \frac{a+2}{a-2} \\ \\(\frac{1}{a}-a) \frac{a^5}{1-a}= \frac{1-a^2}{a}* \frac{a^5}{1-a}=\frac{(1-a)(1+a)}{a}* \frac{a^5}{1-a}=a^4(1+a) \\ \\ (b- \frac{1}{a})*\frac{3a^2}{ab-1}= \frac{ab-1}{a}*\frac{3a^2}{ab-1}=3a \\ \\ (2- \frac{a}{b}):( \frac{a}{2b}-1)= \frac{2b-a}{b}: \frac{a-2b}{2b}=\frac{2b-a}{b}* \frac{2b}{a-2b}= \frac{2b-a}{b}* \frac{2b}{-(2b-a)}=-2$

$\frac{b}{a+b}*( \frac{a^2}{b^2}-1)=\frac{b}{a+b}*\frac{a^2-b^2}{b^2}=\frac{b}{a+b}*\frac{(a-b)(a+b)}{b^2}= \frac{a-b}{b} \\ \\ ( \frac{2x}{y}-2):( \frac{x^2}{y^2}-1)= \frac{2x-2y}{y}: \frac{x^2-y^2}{y^2}=\frac{2(x-y)}{y}* \frac{y^2}{(x-y)(x+y)}= \frac{2y}{x+y} \\ \\ (1+\frac{y}{x})* \frac{x^2}{ax+ay}= \frac{x+y}{x}* \frac{x^2}{a(x+y)}= \frac{x}{a}$

$(3- \frac{3c}{a}):(c^2-a^2)= \frac{3a-3c}{a}* \frac{1}{c^2-a^2}=\frac{3(a-c)}{a}*\frac{1}{(c-a)(c+a)}= \\ =\frac{3(a-c)}{a}*\frac{1}{-(a-c)(c+a)}=- \frac{3}{a(c+a)}$