\(1)\ \frac{7}{2a-4}-\frac{12}{a^2-4}-\frac{3}{a+2}={\frac{7}{ 2(a-2)}}^{\backslash{a\ +\ 2}}-{\frac{12}{ (a-2)(a+2)}}^{\backslash2}-{\frac{3}{a+2}}^{\backslash{ 2(a\ -\ 2)}}=\frac{ 7(a+2)-12\cdot2-3\cdot2(a-2)}{ 2(a-2)(a+2)}=\frac{7a+14-24-6a+12}{ 2(a-2)(a+2)}=\frac{a+2}{ 2(a-2)(a+2)}=\frac{1}{ 2(a-2)}=\frac{1}{2a-4}\)
\(\begin{array}[t]{lcl}{\largeесли}\ a=5,\ {\largeто}&&\frac{1}{2a-4}=\frac{1}{2\cdot5-4}=\frac{1}{10-4}=\frac{1}{6}\end{array}\)
\(2)\ \frac{2c+3}{2c^2-3c}+\frac{2c-3}{2c^2+3c}-\frac{16c}{4c^2-9}={\frac{2c+3}{ c(2c-3)}}^{\backslash{2c\ +\ 3}}+{\frac{2c-3}{ c(2c+3)}}^{\backslash{2c\ -\ 3}}-{\frac{16c}{ (2c-3)(2c+3)}}^{\backslash{c}}=\frac{ (2c+3)(2c+3)+(2c-3)(2c-3)-16c\cdot{c}}{ c(2c-3)(2c+3)}=\frac{ (2c+3)^2+(2c-3)^2-16c^2}{ c(2c-3)(2c+3)}=\frac{4c^2+12c+9+4c^2-12c+9-16c^2}{ c(2c-3)(2c+3)}=\frac{{-}8c^2+18}{ c(2c-3)(2c+3)}=\frac{ {-}(8c^2-18)}{ c(2c-3)(2c+3)}={-}\frac{ 2(4c^2-9)}{ c(4c^2-9)}={-}\frac{2}{c}\)
\(\begin{array}[t]{lcl}{\largeесли}\ c={-}0{,}8,\ {\largeто}&&{-}\frac{2}{c}={-}\frac{2}{{-}0{,}8}=\frac{20}{8}=\frac{5}{2}=2\frac{1}{2}=2{,}5\end{array}\)
\(3)\ \frac{m^2+16n^2}{m^2-16n^2}-\frac{m+4n}{2m-8n}={\frac{m^2+16n^2}{ (m-4n)(m+4n)}}^{\backslash2}-{\frac{m+4n}{ 2(m-4n)}}^{\backslash{m\ +\ 4n}}=\frac{ 2(m^2+16n^2)-(m+4n)(m+4n)}{ 2(m-4n)(m+4n)}=\frac{2m^2+32n^2-(m+4n)^2}{ 2(m-4n)(m+4n)}=\frac{2m^2+32n^2-(m^2+8mn+16n^2)}{ 2(m-4n)(m+4n)}=\frac{2m^2+32n^2-m^2-8mn-16n^2}{ 2(m-4n)(m+4n)}=\frac{m^2-8mn+16n^2}{ 2(m-4n)(m+4n)}=\frac{ (m-4n)^2}{ 2(m-4n)(m+4n)}=\frac{m-4n}{ 2(m+4n)}=\frac{m-4n}{2m+8n}\)
\(\begin{array}[t]{lcl}{\largeесли}\ m=3,\ n=0{,}5,\ {\largeто}&&\frac{m-4n}{2m+8n}=\frac{3-4\cdot0{,}5}{2\cdot3+8\cdot0{,}5}=\frac{3-2}{6+4}=\frac{1}{10}=0{,}1\end{array}\)
