123. Выполните действия: а) \((a^2-x^2)\cdot\left(\frac1{a+x}+\frac1{a^2-x^2}-\frac1{a-x}\right)\); б) \(\frac{by}{b^2-y^2}:\left(\frac1{b^2-y^2}+\frac1{b^2+2by+y^2}\right)\); в) \(\left(x^2+\frac{27}{x}\right)\cdot\left(\frac1{x+3}+\frac1{x^2-3x+9}\right)\); г) \(\left(x-\frac1x\right)\cdot\left(\frac1{x-1}-\frac1{x+1}+5\right)\); д) \(\frac{a+b}{a-b}:\left(\frac{a-b}{a+b}+\frac{a+b}{a-b}\right)\); е) \(\left(\frac1{x-y}-\frac1{x+y}\right)\cdot\left(x-\frac{3x+y}{4}\right)\).

\((a^2-x^2)\cdot\left(\frac1{a+x}+\frac1{a^2-x^2}-\frac1{a-x}\right)=(a-x)+1-(a+x)=1-2x.\)

\(\frac1{b^2-y^2}+\frac1{(b+y)^2}=\frac{2b}{(b-y)(b+y)^2}.\) Тогда \(\frac{by}{(b-y)(b+y)}:\frac{2b}{(b-y)(b+y)^2}=\frac{y(b+y)}2.\)

\(x^2+\frac{27}{x}=\frac{x^3+27}{x}=\frac{(x+3)(x^2-3x+9)}x,\quad \frac1{x+3}+\frac1{x^2-3x+9}=\frac{x^2-2x+12}{(x+3)(x^2-3x+9)}.\) Произведение равно \(\frac{x^2-2x+12}{x}.\)

\(x-\frac1x=\frac{x^2-1}{x},\quad \frac1{x-1}-\frac1{x+1}+5=\frac{5x^2-3}{x^2-1}.\) Поэтому произведение равно \(\frac{5x^2-3}{x}.\)

\(\frac{a-b}{a+b}+\frac{a+b}{a-b}=\frac{2(a^2+b^2)}{a^2-b^2}.\) Следовательно, \(\frac{a+b}{a-b}:\frac{2(a^2+b^2)}{a^2-b^2}=\frac{(a+b)^2}{2(a^2+b^2)}.\)

\(\frac1{x-y}-\frac1{x+y}=\frac{2y}{(x-y)(x+y)},\quad x-\frac{3x+y}{4}=\frac{x-y}{4}.\) Произведение равно \(\frac{y}{2(x+y)}.\)