Задание 494

Решение:

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

• \({\largeб)}\ \frac{1-a^3}{1+a+a^2}=\frac{ (1-a)(1+a+a^2)}{1+a+a^2}=\frac{1-a}{1}=1-a\)

• \({\largeв)}\ \frac{x^3-y^3}{x^2-y^2}=\frac{ (x-y)(x^2+xy+y^2)}{ (x+y)(x-y)}=\frac{x^2+xy+y^2}{x+y}\)

• \({\largeг)}\ \frac{2p^2-2p+2}{p^3+1}=\frac{ 2(p^2-p+1)}{ (p+1)(p^2-p+1)}=\frac{2}{p+1}\)

• \({\largeд)}\ \frac{a^2-4a+4}{a^2-4}=\frac{ (a-2)^2}{ (a+2)(a-2)}=\frac{ (a-2)(a-2)}{ (a+2)(a-2)}=\frac{a-2}{a+2}\)

• \({\largeе)}\ \frac{3x^2+6xy+3y^2}{12y^2-12x^2}=\frac{ 3(x^2+2xy+y^2)}{ 12(y^2-x^2)}=\frac{ 3(x+y)^2}{ 12(y+x)(y-x)}=\frac{ 3(y+x)(y+x)}{ 12(y+x)(y-x)}=\frac{y+x}{ 4(y-x)}=\frac{y+x}{4y-4x}\)

• \({\largeж)}\ \frac{m^2-n^2}{n^3-m^3}=\frac{ (m+n)(m-n)}{ (n-m)(n^2+mn+m^2)}={-}\frac{ (m+n)(m-n)}{ (m-n)(n^2+mn+m^2)}={-}\frac{m+n}{n^2+mn+m^2}\)

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

• \({\largeи)}\ \frac{6a^2-6b^2}{3a^3+3b^3}=\frac{ 6(a^2-b^2)}{ 3(a^3+b^3)}=\frac{ 6(a+b)(a-b)}{ 3(a+b)(a^2-ab+b^2)}=\frac{ 2(a-b)}{a^2-ab+b^2}=\frac{2a-2b}{a^2-ab+b^2}\)

• \({\largeк)}\ \frac{ (x^3-y^3)(x+y)}{x^2-y^2}=\frac{ (x-y)(x^2+xy+y^2)(x+y)}{ (x+y)(x-y)}=\frac{x^2+xy+y^2}{1}=x^2+xy+y^2\)