Задание 521

Решение:

• \({\largeа)}\ \frac{2a}{a^2-9}+\frac{3}{a-3}=\frac{2a^{\backslash1}}{ (a-3)(a+3)\phantom{^{\backslash1}}}+\frac{3^{\backslash{a\ +\ 3}}}{a-3\phantom{^{\backslash{a\ +\ 3}}}}=\frac{2a\cdot1}{ (a-3)(a+3)\cdot1}+\frac{ 3\cdot(a+3)}{ (a-3)\cdot(a+3)}=\frac{2a}{a^2-9}+\frac{3a+9}{a^2-9}=\frac{2a+3a+9}{a^2-9}=\frac{5a+9}{a^2-9}\)

• \({\largeб)}\ \frac{5}{m+n}-\frac{4n}{m^2-n^2}=\frac{5^{\backslash{m\ -\ n}}}{m+n\phantom{^{\backslash{m\ -\ n}}}}-\frac{4n^{\backslash1}}{ (m+n)(m-n)\phantom{^{\backslash1}}}=\frac{ 5\cdot(m-n)}{ (m+n)\cdot(m-n)}-\frac{4n\cdot1}{ (m+n)(m-n)\cdot1}=\frac{5m-5n}{m^2-n^2}-\frac{4n}{ m^2-n^2}=\frac{5m-5n-4n}{m^2-n^2}=\frac{5m-9n}{m^2-n^2}\)

• \({\largeв)}\ \frac{x}{4-9x^2}+\frac{1}{3x+2}=\frac{x^{\backslash1}}{ (2+3x)(2-3x)\phantom{^{\backslash1}}}+\frac{1^{\backslash{2\ -\ 3x}}}{2+3x\phantom{^{\backslash{2\ -\ 3x}}}}=\frac{x\cdot1}{ (2+3x)(2-3x)\cdot1}+\frac{ 1\cdot(2-3x)}{ (2+3x)\cdot(2-3x)}=\frac{x}{4-9x^2}+\frac{2-3x}{4-9x^2}=\frac{x+2-3x}{4-9x^2}=\frac{2-2x}{4-9x^2}\)

• \({\largeг)}\ \frac{1}{2p+4q}-\frac{q}{4q^2-p^2}=\frac{1^{\backslash{2q\ -\ p}}}{ 2(p+2q)\phantom{^{\backslash{2q\ -\ p}}}}-\frac{q^{\backslash2}}{ (2q+p)(2q-p)\phantom{^{\backslash2}}}=\frac{ 1\cdot(2q-p)}{ 2(2q+p)\cdot(2q-p)}-\frac{q\cdot2}{ (2q+p)(2q-p)\cdot2}=\frac{2q-p}{ 2(4q^2-p^2)}-\frac{2q}{ 2(4q^2-p^2)}=\frac{2q-p-2q}{ 2(4q^2-p^2)}=\frac{{-}p}{ 2(4q^2-p^2)}={-}\frac{p}{ 2(4q^2-p^2)}\)

• \({\largeд)}\ \frac{1}{a^2+ab+b^2}+\frac{b}{a^3-b^3}=\frac{1^{\backslash{a\ -\ b}}}{a^2+ab+b^2\phantom{^{\backslash{a\ -\ b}}}}+\frac{b^{\backslash1}}{ (a-b)(a^2+ab+b^2)\phantom{^{\backslash1}}}=\frac{ 1\cdot(a-b)}{ (a^2+ab+b^2)\cdot(a-b)}+\frac{b\cdot1}{ (a-b)(a^2+ab+b^2)\cdot1}=\frac{a-b}{a^3-b^3}+\frac{b}{a^3-b^3}=\frac{a-b+b}{a^3-b^3}=\frac{a}{a^3-b^3}\)

• \({\largeе)}\ \frac{m^2+n^2}{m^3+n^3}-\frac{1}{ 2(m+n)}=\frac{m^2+n^{2\backslash2}}{ (m+n)(m^2-mn+n^2)\phantom{^{\backslash2}}}-\frac{1^{\backslash{m^2\ -\ mn\ +\ n^2}}}{ 2(m+n)\phantom{^{\backslash{m^2\ -\ mn\ +\ n^2}}}}=\frac{ (m^2+n^2)\cdot2}{ (m+n)(m^2-mn+n^2)\cdot2}-\frac{ 1\cdot(m^2-mn+n^2)}{ 2(m+n)\cdot(m^2-mn+n^2)}=\frac{ 2m^2+2n^2}{ 2(m^3+n^3)}-\frac{m^2-mn+n^2}{ 2(m^3+n^3)}=\frac{ 2m^2+2n^2-(m^2-mn+n^2)}{ 2(m^3+n^3)}=\frac{2m^2+2n^2-m^2+mn-n^2}{ 2(m^3+n^3)}=\frac{m^2+mn+n^2}{ 2(m^3+n^3)}\)

• \({\largeж)}\ \frac{x^2-2xy}{ (x-2y)^3}+\frac{1}{2y-x}=\frac{ x(x-2y)}{ (x-2y)^3}-\frac{1}{x-2y}=\frac{x^{\backslash1}}{ (x-2y)^2\phantom{^{\backslash1}}}-\frac{1^{\backslash{x\ -\ 2y}}}{x-2y\phantom{^{\backslash{x\ -\ 2y}}}}=\frac{x\cdot1}{ (x-2y)^2\cdot1}-\frac{ 1\cdot(x-2y)}{ (x-2y)\cdot(x-2y)}=\frac{x}{ (x-2y)^2}-\frac{x-2y}{ (x-2y)^2}=\frac{ x-(x-2y)}{ (x-2y)^2}=\frac{x-x+2y}{ (x-2y)^2}=\frac{2y}{ (x-2y)^2}\)

• \({\largeз)}\ \frac{ 2(p+q)}{p^3-q^3}+\frac{3}{q^2-p^2}=\frac{ 2(p+q)^{\backslash{p\ +\ q}}}{ (p-q)(p^2+pq+q^2)}-\frac{3^{\backslash{p^2\ +\ pq\ +\ q^2}}}{ (p-q)(p+q)}=\frac{ 2(p+q)\cdot(p+q)}{ (p-q)(p^2+pq+q^2)\cdot(p+q)}-\frac{ 3\cdot(p^2+pq+q^2)}{ (p-q)(p+q)\cdot(p^2+pq+q^2)}=\frac{ 2(p+q)^2}{ (p^3-q^3)(p+q)}-\frac{3p^2+3pq+3q^2}{ (p^3-q^3)(p+q)}=\frac{ 2(p^2+2pq+q^2)-(3p^2+3pq+3q^2)}{ (p^3-q^3)(p+q)}=\frac{2p^2+4pq+2q^2-3p^2-3pq-3q^2}{ (p^3-q^3)(p+q)}=\frac{{-}p^2+pq-q^2}{ (p^3-q^3)(p+q)}={-}\frac{p^2-pq+q^2}{ (p^3-q^3)(p+q)}\)