Задание 515

Решение:

• \({\largeа)}\ \frac{1^{\backslash{a\ +\ b}}}{a\phantom{^{\backslash{a\ +\ b}}}}+\frac{1^{\backslash{a}}}{a+b\phantom{^{\backslash{a}}}}=\frac{ 1\cdot(a+b)}{ a\cdot(a+b)}+\frac{1\cdot{a}}{ (a+b)\cdot{a}}=\frac{a+b}{ a(a+b)}+\frac{a}{ a(a+b)}=\frac{a+b+a}{ a(a+b)}=\frac{2a+b}{ a(a+b)}\)

• \({\largeб)}\ \frac{1^{\backslash{a\ -\ b}}}{a+b\phantom{^{\backslash{a\ -\ b}}}}+\frac{1^{\backslash{a\ +\ b}}}{a-b\phantom{^{\backslash{a\ +\ b}}}}=\frac{ 1\cdot(a-b)}{ (a+b)\cdot(a-b)}+\frac{ 1\cdot(a+b)}{ (a-b)\cdot(a+b)}=\frac{a-b}{a^2-b^2}+\frac{a+b}{a^2-b^2}=\frac{a-b+a+b}{a^2-b^2}=\frac{2a}{a^2-b^2}\)

• \({\largeв)}\ \frac{1^{\backslash{n}}}{m+n\phantom{^{\backslash{n}}}}-\frac{1^{\backslash{m\ +\ n}}}{n\phantom{^{\backslash{m\ +\ n}}}}=\frac{1\cdot{n}}{ (m+n)\cdot{n}}-\frac{ 1\cdot(m+n)}{ n\cdot(m+n)}=\frac{n}{ n(m+n)}-\frac{m+n}{ n(m+n)}=\frac{ n-(m+n)}{ n(m+n)}=\frac{n-m-n}{ n(m+n)}=\frac{{-}m}{ n(m+n)}={-}\frac{m}{ n(m+n)}\)

• \({\largeг)}\ \frac{1^{\backslash{x\ +\ y}}}{x-y\phantom{^{\backslash{x\ +\ y}}}}-\frac{1^{\backslash{x\ -\ y}}}{x+y\phantom{^{\backslash{x\ -\ y}}}}=\frac{ 1\cdot(x+y)}{ (x-y)\cdot(x+y)}-\frac{ 1\cdot(x-y)}{ (x+y)\cdot(x-y)}=\frac{x+y}{x^2-y^2}-\frac{x-y}{x^2-y^2}=\frac{ x+y-(x-y)}{x^2-y^2}=\frac{x+y-x+y}{x^2-y^2}=\frac{2y}{x^2-y^2}\)

• \({\largeд)}\ \frac{2^{\backslash{a\ +\ b}}}{a-b\phantom{^{\backslash{a\ +\ b}}}}+\frac{3^{\backslash{a\ -\ b}}}{a+b\phantom{^{\backslash{a\ -\ b}}}}=\frac{ 2\cdot(a+b)}{ (a-b)\cdot(a+b)}+\frac{ 3\cdot(a-b)}{ (a+b)\cdot(a-b)}=\frac{2a+2b}{a^2-b^2}+\frac{3a-3b}{a^2-b^2}=\frac{2a+2b+3a-3b}{a^2-b^2}=\frac{5a-b}{a^2-b^2}\)

• \({\largeе)}\ \frac{4^{\backslash{p\ +\ q}}}{p-q\phantom{^{\backslash{p\ +\ q}}}}-\frac{3^{\backslash{p\ -\ q}}}{p+q\phantom{^{\backslash{p\ -\ q}}}}=\frac{ 4\cdot(p+q)}{ (p-q)\cdot(p+q)}-\frac{ 3\cdot(p-q)}{ (p+q)\cdot(p-q)}=\frac{4p+4q}{p^2-q^2}-\frac{3p-3q}{p^2-q^2}=\frac{ 4p+4q-(3p-3q)}{p^2-q^2}=\frac{4p+4q-3p+3q}{p^2-q^2}=\frac{p+7q}{p^2-q^2}\)

• \({\largeж)}\ \frac{2a^{\backslash{a\ +\ b}}}{a-2b\phantom{^{\backslash{a\ +\ b}}}}+\frac{3a^{\backslash{a\ -\ 2b}}}{a+b\phantom{^{\backslash{a\ -\ 2b}}}}=\frac{ 2a\cdot(a+b)}{ (a-2b)\cdot(a+b)}+\frac{ 3a\cdot(a-2b)}{ (a+b)\cdot(a-2b)}=\frac{2a^2+2ab}{ (a-2b)(a+b)}+\frac{3a^2-6ab}{ (a-2b)(a+b)}=\frac{2a^2+2ab+3a^2-6ab}{ (a-2b)(a+b)}=\frac{5a^2-4ab}{ (a-2b)(a+b)}\)

• \({\largeз)}\ \frac{3x^{\backslash{2x\ -\ y}}}{x-y\phantom{^{\backslash{2x\ -\ y}}}}-\frac{2x^{\backslash{x\ -\ y}}}{2x-y\phantom{^{\backslash{x\ -\ y}}}}=\frac{ 3x\cdot(2x-y)}{ (x-y)\cdot(2x-y)}-\frac{ 2x\cdot(x-y)}{ (2x-y)\cdot(x-y)}=\frac{6x^2-3xy}{ (x-y)(2x-y)}-\frac{2x^2-2xy}{ (x-y)(2x-y)}=\frac{ 6x^2-3xy-(2x^2-2xy)}{ (x-y)(2x-y)}=\frac{6x^2-3xy-2x^2+2xy}{ (x-y)(2x-y)}=\frac{4x^2-xy}{ (x-y)(2x-y)}\)

• \({\largeи)}\ \frac{5m}{2m-n}-\frac{3m}{n-m}=\frac{5m^{\backslash{m\ -\ n}}}{2m-n\phantom{^{\backslash{m\ -\ n}}}}+\frac{3m^{\backslash{2m\ -\ n}}}{m-n\phantom{^{\backslash{2m\ -\ n}}}}=\frac{ 5m\cdot(m-n)}{ (2m-n)\cdot(m-n)}+\frac{ 3m\cdot(2m-n)}{ (m-n)\cdot(2m-n)}=\frac{5m^2-5mn}{ (2m-n)(m-n)}+\frac{6m^2-3mn}{ (2m-n)(m-n)}=\frac{5m^2-5mn+6m^2-3mn}{ (2m-n)(m-n)}=\frac{11m^2-8mn}{ (2m-n)(m-n)}\)

• \({\largeк)}\ \frac{4p}{q-2p}-\frac{2p}{2p+q}=\frac{4p^{\backslash{q\ +\ 2p}}}{q-2p\phantom{^{\backslash{q\ +\ 2p}}}}-\frac{2p^{\backslash{q\ -\ 2p}}}{q+2p\phantom{^{\backslash{q\ -\ 2p}}}}=\frac{ 4p\cdot(q+2p)}{ (q-2p)\cdot(q+2p)}-\frac{ 2p\cdot(q-2p)}{ (q+2p)\cdot(q-2p)}=\frac{4pq+8p^2}{q^2-4p^2}-\frac{2pq-4p^2}{q^2-4p^2}=\frac{ 4pq+8p^2-(2pq-4p^2)}{q^2-4p^2}=\frac{4pq+8p^2-2pq+4p^2}{q^2-4p^2}=\frac{12p^2+2pq}{q^2-4p^2}\)

• \({\largeл)}\ \frac{7}{2x-y}-\frac{5}{y-2x}=\frac{7}{2x-y}+\frac{5}{2x-y}=\frac{7+5}{2x-y}=\frac{12}{2x-y}\)

• \({\largeм)}\ \frac{5x}{x-3y}+\frac{4x+3y}{3y-x}=\frac{5x}{x-3y}-\frac{4x+3y}{x-3y}=\frac{ 5x-(4x+3y)}{x-3y}=\frac{ 5x-4x-3y}{x-3y}=\frac{x-3y}{x-3y}=1\)