Задание 528

Решение:

• \({\largeа)}\ \frac{m^3+n^3}{2m}\cdot\frac{4mn}{m^2-mn+n^2}=\frac{ (m+n)(m^2-mn+n^2)}{2m}\cdot\frac{4mn}{m^2-mn+n^2}=\frac{m+n}{1}\cdot\frac{2n}{1}=\frac{2mn+2n^2}{1}=2mn+2n^2\)

• \({\largeб)}\ \frac{2a}{a^3-b^3}:\frac{6ab}{a^2-b^2}=\frac{2a}{a^3-b^3}\cdot\frac{a^2-b^2}{6ab}=\frac{2a}{ (a-b)(a^2+ab+b^2)}\cdot\frac{ (a+b)(a-b)}{6ab}=\frac{1}{a^2+ab+b^2}\cdot\frac{a+b}{3b}=\frac{a+b}{ 3b(a^2+ab+b^2)}=\frac{a+b}{3a^2b+3ab^2+3b^3}\)

• \({\largeв)}\ \frac{m^3-n^3}{m^3+n^3}:\frac{ (m-n)^2}{m^2-n^2}=\frac{m^3-n^3}{m^3+n^3}\cdot\frac{m^2-n^2}{ (m-n)^2}=\frac{ (m-n)(m^2+mn+n^2)}{ (m+n)(m^2-mn+n^2)}\cdot\frac{ (m+n)(m-n)}{ (m-n)^2}=\frac{m^2+mn+n^2}{m^2-mn+n^2}\cdot\frac{1}{1}=\frac{m^2+mn+n^2}{m^2-mn+n^2}\)

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

• \({\largeд)}\ \frac{p^2-4q^2}{ (p+2q)^2}:\frac{p^3-8q^3}{4q^2+2pq+p^2}=\frac{p^2-4q^2}{ (p+2q)^2}\cdot\frac{4q^2+2pq+p^2}{p^3-8q^3}=\frac{ (p+2q)(p-2q)}{ (p+2q)^2}\cdot\frac{4q^2+2pq+p^2}{ (p-2q)(p^2+2pq+4q^2)}=\frac{1}{p+2q}\cdot\frac{1}{1}=\frac{1}{p+2q}\)

• \({\largeе)}\ \frac{12a^2+6ab}{8a^3-b^3}\cdot\frac{4a^2+2ab+b^2}{3a^2-6ab}=\frac{ 6a(2a+b)}{ (2a-b)(4a^2+2ab+b^2)}\cdot\frac{4a^2+2ab+b^2}{ 3a(a-2b)}=\frac{ 2(2a+b)}{2a-b}\cdot\frac{1}{a-2b}=\frac{4a+2b}{ (2a-b)(a-2b)}=\frac{4a+2b}{2a^2-4ab-ab+2b^2}=\frac{4a+2b}{2a^2-5ab+2b^2}\)