Упражнение 124

\(1)\ \frac{4b}{a^2-b^2}+\frac{a-b}{a^2+ab}+\frac{a+b}{b^2-ab}=\frac{4b}{a^2-b^2}+\frac{a-b}{a^2+ab}+\frac{a+b}{ {-}(ab-b^2)}={\frac{4b}{ (a-b)(a+b)}}^{\backslash{ab}}+{\frac{a-b}{ a(a+b)}}^{\backslash{ b(a\ -\ b)}}-{\frac{a+b}{ b(a-b)}}^{\backslash{ a(a\ +\ b)}}=\frac{4b\cdot{ab}+b(a-b)(a-b)-a(a+b)(a+b)}{ ab(a-b)(a+b)}=\frac{4ab^2+b(a-b)^2-a(a+b)^2}{ ab(a-b)(a+b)}=\frac{4ab^2+b(a^2-2ab+b^2)-a(a^2+2ab+b^2)}{ ab(a-b)(a+b)}=\frac{4ab^2+a^2b-2ab^2+b^3-a^3-2a^2b-ab^2}{ ab(a-b)(a+b)}=\frac{ab^2-a^2b+b^3-a^3}{ ab(a-b)(a+b)}=\frac{ (ab^2-a^2b)+(b^3-a^3)}{ ab(a-b)(a+b)}=\frac{ ab(b-a)+(b-a)(b^2+ab+a^2)}{ ab(a-b)(a+b)}=\frac{ (b-a)(ab+b^2+ab+a^2)}{ ab(a-b)(a+b)}=\frac{ {-}(a-b)(a^2+2ab+b^2)}{ ab(a-b)(a+b)}={-}\frac{ (a-b)(a+b)^2}{ ab(a-b)(a+b)}={-}\frac{a+b}{ab};\)

\(2)\ \frac{1}{x-2}+\frac{1}{x+2}-\frac{x}{x^2-4}+\frac{x^2+4}{8x-2x^3}=\frac{1}{x-2}+\frac{1}{x+2}-\frac{x}{x^2-4}+\frac{x^2+4}{ {-}(2x^3-8x)}=\frac{1}{x-2}+\frac{1}{x+2}-\frac{x}{ (x-2)(x+2)}-\frac{x^2+4}{ 2x(x^2-4)}={\frac{1}{x-2}}^{\backslash{ 2x(x\ +\ 2)}}+{\frac{1}{x+2}}^{\backslash{ 2x(x\ -\ 2)}}-{\frac{x}{ (x-2)(x+2)}}^{\backslash2x}-{\frac{x^2+4}{ 2x(x-2)(x+2)}}^{\backslash1}=\frac{ 2x(x+2)+2x(x-2)-x\cdot2x-(x^2+4)}{ 2x(x-2)(x+2)}=\frac{2x^2+4x+2x^2-4x-2x^2-x^2-4}{ 2x(x-2)(x+2)}=\frac{x^2-4}{ 2x(x-2)(x+2)}=\frac{ (x-2)(x+2)}{ 2x(x-2)(x+2)}=\frac{1}{2x};\)

\(3)\ \frac{1}{ (a-5b)^2}-\frac{2}{a^2-25b^2}+\frac{1}{ (a+5b)^2}={\frac{1}{ (a-5b)^2}}^{\backslash{ (a\ +\ 5b)^2}}-{\frac{2}{ (a-5b)(a+5b)}}^{\backslash{ (a\ -\ 5b)(a\ +\ 5b)}}+{\frac{1}{ (a+5b)^2}}^{\backslash{ (a\ -\ 5b)^2}}=\frac{ (a+5b)^2-2(a-5b)(a+5b)+(a-5b)^2}{ (a-5b)^2(a+5b)^2}=\frac{ (a+5b)^2-2(a^2-25b^2)+(a-5b)^2}{ (a-5b)^2(a+5b)^2}=\frac{a^2+10ab+25b^2-2a^2+50b^2+a^2-10ab+25b^2}{ (a-5b)^2(a+5b)^2}=\frac{100b^2}{ (a-5b)^2(a+5b)^2}=\frac{100b^2}{ (a^2-25b^2)^2};\)

\(4)\ \frac{x^2+9x+18}{xy+3y-2x-6}-\frac{x+5}{y-2}=\frac{x^2+9x+18}{ (xy+3y)-(2x+6)}-\frac{x+5}{y-2}=\frac{x^2+9x+18}{ y(x+3)-2(x+3)}-\frac{x+5}{y-2}={\frac{x^2+9x+18}{ (x+3)(y-2)}}^{\backslash1}-{\frac{x+5}{y-2}}^{\backslash{x\ +\ 3}}=\frac{x^2+9x+18-(x+5)(x+3)}{ (x+3)(y-2)}=\frac{x^2+9x+18-(x^2+3x+5x+15)}{ (x+3)(y-2)}=\frac{x^2+9x+18-x^2-3x-5x-15}{ (x+3)(y-2)}=\frac{x+3}{ (x+3)(y-2)}=\frac{1}{y-2}.\)