\(1)\ \frac{a^2+1}{a^2-2a+1}+\frac{a+1}{a-1}={\frac{a^2+1}{ (a-1)^2}}^{\backslash1}+{\frac{a+1}{a-1}}^{\backslash{a\ -\ 1}}=\frac{a^2+1+(a+1)(a-1)}{ (a-1)^2}=\frac{a^2+1+a^2-1}{ (a-1)^2}=\frac{2a^2}{ (a-1)^2}=\frac{2a^2}{a^2-2a+1};\)
\(2)\ \frac{a^2+b^2}{a^2-b^2}-\frac{a-b}{a+b}={\frac{a^2+b^2}{ (a-b)(a+b)}}^{\backslash1}-{\frac{a-b}{a+b}}^{\backslash{a\ -\ b}}=\frac{a^2+b^2-(a-b)(a-b)}{ (a-b)(a+b)}=\frac{a^2+b^2-(a-b)^2}{ (a-b)(a+b)}=\frac{a^2+b^2-(a^2-2ab+b^2)}{ (a-b)(a+b)}=\frac{a^2+b^2-a^2+2ab-b^2}{ (a-b)(a+b)}=\frac{2ab}{ (a-b)(a+b)}=\frac{2ab}{a^2-b^2};\)
\(3)\ \frac{c+7}{c-7}+\frac{28c}{49-c^2}=\frac{c+7}{c-7}+\frac{28c}{ {-}(c^2-49)}={\frac{c+7}{c-7}}^{\backslash{c\ +\ 7}}-{\frac{28c}{ (c-7)(c+7)}}^{\backslash1}=\frac{ (c+7)(c+7)-28c}{ (c-7)(c+7)}=\frac{ (c+7)^2-28c}{ (c-7)(c+7)}=\frac{c^2+14c+49-28c}{ (c-7)(c+7)}=\frac{c^2-14c+49}{ (c-7)(c+7)}=\frac{ (c-7)^2}{ (c-7)(c+7)}=\frac{c-7}{c+7};\)
\(4)\ \frac{5a+3}{2a^2+6a}+\frac{6-3a}{a^2-9}={\frac{5a+3}{ 2a(a+3)}}^{\backslash{a\ -\ 3}}+{\frac{6-3a}{ (a-3)(a+3)}}^{\backslash2a}=\frac{ (5a+3)(a-3)+2a(6-3a)}{ 2a(a-3)(a+3)}=\frac{5a^2-15a+3a-9+12a-6a^2}{ 2a(a-3)(a+3)}=\frac{{-}a^2-9}{ 2a(a-3)(a+3)}=\frac{ {-}(a^2+9)}{ 2a(a^2-9)}={-}\frac{a^2+9}{2a^3-18a};\)
\(5)\ \frac{a}{a^2-4a+4}-\frac{a+4}{a^2-4}={\frac{a}{ (a-2)^2}}^{\backslash{a\ +\ 2}}-{\frac{a+4}{ (a-2)(a+2)}}^{\backslash{a\ -\ 2}}=\frac{ a(a+2)-(a+4)(a-2)}{ (a-2)^2(a+2)}=\frac{a^2+2a-(a^2-2a+4a-8)}{ (a-2)^2(a+2)}=\frac{a^2+2a-a^2+2a-4a+8}{ (a-2)^2(a+2)}=\frac{8}{ (a-2)^2(a+2)};\)
\(6)\ \frac{2p}{p-5}-\frac{5}{p+5}+\frac{2p^2}{25-p^2}=\frac{2p}{p-5}-\frac{5}{p+5}+\frac{2p^2}{ {-}(p^2-25)}={\frac{2p}{p-5}}^{\backslash{p\ +\ 5}}-{\frac{5}{p+5}}^{\backslash{p\ -\ 5}}-{\frac{2p^2}{ (p-5)(p+5)}}^{\backslash1}=\frac{ 2p(p+5)-5(p-5)-2p^2}{ (p-5)(p+5)}=\frac{2p^2+10p-5p+25-2p^2}{ (p-5)(p+5)}=\frac{5p+25}{ (p-5)(p+5)}=\frac{ 5(p+5)}{ (p-5)(p+5)}=\frac{5}{p-5};\)
\(7)\ \frac{1}{y}-\frac{y+8}{16-y^2}-\frac{2}{y-4}=\frac{1}{y}-\frac{y+8}{16-y^2}-\frac{2}{ {-}(4-y)}={\frac{1}{y}}^{\backslash{ (4-y)(4+y)}}-{\frac{y+8}{ (4-y)(4+y)}}^{\backslash{y}}+{\frac{2}{4-y}}^{\backslash{ y(4+y)}}=\frac{ (4-y)(4+y)-y(y+8)+2y(4+y)}{ y(4-y)(4+y)}=\frac{16-y^2-y^2-8y+8y+2y^2}{ y(4-y)(4+y)}=\frac{16}{ y(4-y)(4+y)}=\frac{16}{ y(16-y^2)}=\frac{16}{16y-y^3};\)
\(8)\ \frac{2b-1}{4b+2}+\frac{4b}{4b^2-1}+\frac{2b+1}{3-6b}={\frac{2b-1}{ 2(2b+1)}}^{\backslash{ 3(2b\ -\ 1)}}+{\frac{4b}{ (2b-1)(2b+1)}}^{\backslash6}+{\frac{2b+1}{ {-}3(2b-1)}}^{\backslash{ 2(2b\ +\ 1)}}=\frac{ 3(2b-1)(2b-1)+4b\cdot6-2(2b+1)(2b+1)}{ 6(2b-1)(2b+1)}=\frac{ 3(2b-1)^2+24b-2(2b+1)^2}{ 6(2b-1)(2b+1)}=\frac{ 3(4b^2-4b+1)+24b-2(4b^2+4b+1)}{ 6(2b-1)(2b+1)}=\frac{12b^2-12b+3+24b-8b^2-8b-2}{ 6(2b-1)(2b+1)}=\frac{4b^2+4b+1}{ 6(2b-1)(2b+1)}=\frac{ (2b+1)^2}{ 6(2b-1)(2b+1)}=\frac{2b+1}{ 6(2b-1)}=\frac{2b+1}{12b-6}.\)
