\(1)\ \left(\frac{a^2}{b^3-ab^2}+\frac{a-b}{b^2}-\frac{1}{b}\right):\left(\frac{a+b}{b-a}-\frac{b-a}{a+b}+\frac{6a^2}{a^2-b^2}\right)={-}\frac{a+b}{2ab}\)
\(1)\ \frac{a^2}{b^3-ab^2}+\frac{a-b}{b^2}-\frac{1}{b}={\frac{a^2}{ b^2(b-a)}}^{\backslash1}+{\frac{a-b}{b^2}}^{\backslash{b\ -\ a}}-{\frac{1}{b}}^{\backslash{b}(b\ -\ a)}=\frac{a^2+(a-b)(b-a)-b(b-a)}{ b^2(b-a)}=\frac{a^2+ab-a^2-b^2+ab-b^2+ab}{ b^2(b-a)}=\frac{3ab-2b^2}{ b^2(b-a)}=\frac{ b(3a-2b)}{ b^2(b-a)}=\frac{3a-2b}{ b(b-a)}\)
\(2)\ \frac{a+b}{b-a}-\frac{b-a}{a+b}+\frac{6a^2}{a^2-b^2}={\frac{a+b}{ {-}(a-b)}}^{\backslash{a\ +\ b}}-{\frac{b-a}{a+b}}^{\backslash{a\ -\ b}}+{\frac{6a^2}{ (a-b)(a+b)}}^{\backslash1}=\frac{ {-}(a+b)^2-(b-a)(a-b)+6a^2}{ (a-b)(a+b)}=\frac{ {-}(a^2+2ab+b^2)-(ab-b^2-a^2+ab)+6a^2}{ (a-b)(a+b)}=\frac{ {-}a^2-2ab-b^2-ab+b^2+a^2-ab+6a^2}{ (a-b)(a+b)}=\frac{6a^2-4ab}{ (a-b)(a+b)}=\frac{ 2a(3a-2b)}{ (a-b)(a+b)}\)
\(3)\ \frac{3a-2b}{ b(b-a)}:\frac{ 2a(3a-2b)}{ (a-b)(a+b)}=\frac{3a-2b}{ {-}b(a-b)}\cdot\frac{ (a-b)(a+b)}{ 2a(3a-2b)}={-}\frac{a+b}{2ab}\)
\(2)\ \left(\frac{a+2}{4a^3-4a^2+a}-\frac{2-a}{1-8a^3}\cdot\frac{4a^2+2a+1}{2a^2+a}\right):\left(\frac{1}{1-2a}\right)^2-\frac{8a-1}{2a^2+a}=\frac{1}{a}\)
\(1)\ \frac{2-a}{1-8a^3}\cdot\frac{4a^2+2a+1}{2a^2+a}=\frac{2-a}{ {-}(8a^3-1)}\cdot\frac{4a^2+2a+1}{ a(2a+1)}=\frac{ {-}(2-a)}{ (2a-1)(4a^2+2a+1)}\cdot\frac{4a^2+2a+1}{ a(2a+1)}=\frac{a-2}{ a(2a-1)(2a+1)}\)
\(2)\ \frac{a+2}{4a^3-4a^2+a}-\frac{a-2}{ a(2a-1)(2a+1)}=\frac{a+2}{ a(4a^2-4a+1)}-\frac{a-2}{ a(2a-1)(2a+1)}={\frac{a+2}{ a(2a-1)^2}}^{\backslash{2a\ +\ 1}}-{\frac{a-2}{ a(2a-1)(2a+1)}}^{\backslash{2a\ -\ 1}}=\frac{ (a+2)(2a+1)-(a-2)(2a-1)}{ a(2a-1)^2(2a+1)}=\frac{2a^2+a+4a+2-(2a^2-a-4a+2)}{ a(2a-1)^2(2a+1)}=\frac{2a^2+a+4a+2-2a^2+a+4a-2}{ a(2a-1)^2(2a+1)}=\frac{10a}{ a(2a-1)^2(2a+1)}=\frac{10}{ (2a-1)^2(2a+1)}\)
\(3)\ \frac{10}{ (2a-1)^2(2a+1)}:\left(\frac{1}{1-2a}\right)^2=\frac{10}{ (2a-1)^2(2a+1)}:\frac{1^2}{ (1-2a)^2}=\frac{10}{ (2a-1)^2(2a+1)}\cdot\frac{ (2a-1)^2}{1}=\frac{10}{2a+1}\)
\(4)\ \frac{10}{2a+1}-\frac{8a-1}{2a^2+a}={\frac{10}{2a+1}}^{\backslash{a}}-{\frac{8a-1}{ a(2a+1)}}^{\backslash1}=\frac{10a-(8a-1)}{ a(2a+1)}=\frac{10a-8a+1}{ a(2a+1)}=\frac{2a+1}{ a(2a+1)}=\frac{1}{a}\)
