Упражнение 107

\(1)\ \frac{b}{b-5}-\frac{4b-1}{4b-20}={\frac{b}{b-5}}^{\backslash4}-{\frac{4b-1}{ 4(b-5)}}^{\backslash1}=\frac{4b-(4b-1)}{ 4(b-5)}=\frac{4b-4b+1}{ 4(b-5)}=\frac{1}{ 4(b-5)}=\frac{1}{4b-20};\)

\(2)\ \frac{2}{m}-\frac{16}{m^2+8m}={\frac{2}{m}}^{\backslash{m\ +\ 8}}-{\frac{16}{ m(m+8)}}^{\backslash1}=\frac{ 2(m+8)-16}{ m(m+8)}=\frac{2m+16-16}{ m(m+8)}=\frac{2m}{ m(m+8)}=\frac{2}{m+8};\)

\(3)\ \frac{a-2}{2a-6}-\frac{a-1}{3a-9}={\frac{a-2}{ 2(a-3)}}^{\backslash3}-{\frac{a-1}{ 3(a-3)}}^{\backslash2}=\frac{ 3(a-2)-2(a-1)}{ 6(a-3)}=\frac{3a-6-2a+2}{ 6(a-3)}=\frac{a-4}{ 6(a-3)}=\frac{a-4}{6a-18};\)

\(4)\ \frac{a^2+b^2}{2a^2+2ab}+\frac{b}{a+b}={\frac{a^2+b^2}{ 2a(a+b)}}^{\backslash1}+{\frac{b}{a+b}}^{\backslash2a}=\frac{a^2+b^2+2ab}{ 2a(a+b)}=\frac{ (a+b)^2}{ 2a(a+b)}=\frac{a+b}{2a};\)

\(5)\ \frac{b+4}{ab-b^2}-\frac{a+4}{a^2-ab}={\frac{b+4}{ b(a-b)}}^{\backslash{a}}-{\frac{a+4}{ a(a-b)}}^{\backslash{b}}=\frac{ a(b+4)-b(a+4)}{ ab(a-b)}=\frac{ab+4a-ab-4b}{ ab(a-b)}=\frac{4a-4b}{ ab(a-b)}=\frac{ 4(a-b)}{ ab(a-b)}=\frac{4}{ab};\)

\(6)\ \frac{c-4}{4c+24}+\frac{4c+9}{c^2+6c}={\frac{c-4}{ 4(c+6)}}^{\backslash{c}}+{\frac{4c+9}{ c(c+6)}}^{\backslash4}=\frac{ c(c-4)+4(4c+9)}{ 4c(c+6)}=\frac{c^2-4c+16c+36}{ 4c(c+6)}=\frac{c^2+12c+36}{ 4c(c+6)}=\frac{ (c+6)^2}{ 4c(c+6)}=\frac{c+6}{4c}.\)