Упражнение 102

\(1)\ {\frac{2}{x}}^{\backslash{x\ +\ 1}}+{\frac{3x-2}{x+1}}^{\backslash{x}}=\frac{ 2(x+1)+x(3x-2)}{ x(x+1)}=\frac{2x+2+3x^2-2x}{ x(x+1)}=\frac{3x^2+2}{ x(x+1)}=\frac{3x^2+2}{x^2+x};\)

\(2)\ {\frac{m}{n}}^{\backslash{m\ +\ n}}-{\frac{m}{m+n}}^{\backslash{n}}=\frac{ m(m+n)-mn}{ n(m+n)}=\frac{m^2+mn-mn}{ n(m+n)}=\frac{m^2}{ n(m+n)}=\frac{m^2}{mn+n^2};\)

\(3)\ {\frac{a}{a-3}}^{\backslash{a\ +\ 3}}-{\frac{3}{a+3}}^{\backslash{a\ -\ 3}}=\frac{ a(a+3)-3(a-3)}{ (a-3)(a+3)}=\frac{a^2+3a-3a+9}{ (a-3)(a+3)}=\frac{a^2+9}{ (a-3)(a+3)}=\frac{a^2+9}{a^2-9};\)

\(4)\ {\frac{c}{3c-1}}^{\backslash{3c\ +\ 1}}-{\frac{c}{3c+1}}^{\backslash{3c\ -\ 1}}=\frac{ c(3c+1)-c(3c-1)}{ (3c-1)(3c+1)}=\frac{3c^2+c-3c^2+c}{ (3c-1)(3c+1)}=\frac{2c}{ (3c-1)(3c+1)}=\frac{2c}{9c^2-1};\)

\(5)\ {\frac{x}{2y+1}}^{\backslash{3y\ -\ 2}}-{\frac{x}{3y-2}}^{\backslash{2y\ +\ 1}}=\frac{ x(3y-2)-x(2y+1)}{ (2y+1)(3y-2)}=\frac{3xy-2x-2xy-x}{ (2y+1)(3y-2)}=\frac{xy-3x}{ (2y+1)(3y-2)}=\frac{xy-3x}{6y^2-4y+3y-2}=\frac{xy-3x}{6y^2-y-2};\)

\(6)\ {\frac{a-b}{b}}^{\backslash{a\ +\ b}}-{\frac{a-b}{a+b}}^{\backslash{b}}=\frac{ (a-b)(a+b)-b(a-b)}{ b(a+b)}=\frac{a^2-b^2-ab+b^2}{ b(a+b)}=\frac{a^2-ab}{ b(a+b)}=\frac{a^2-ab}{ab+b^2}.\)