Question

Не могу решить, пожалуйста помогите! Заранее спасибо!

Answer 1

$\frac{mx^{2}-my^{2}}{2m+8}*\frac{3m+12}{my+mx}=\frac{m(x^{2}-y^{2})}{2(m+4)}*\frac{3(m+4)}{m(x+y)}=\frac{3(x^{2}-y^{2})}{2(x+y)}=\frac{3(x+y)(x-y)}{2(x+y)}=\frac{3(x-y)}{2} \\\\\frac{ax+ay}{x^{2}-2xy+y^{2}}*\frac{x^{2}-xy}{7x+7y}=\frac{a(x+y)}{(x-y)^{2}}*\frac{x(x-y)}{7(x+y)}=\frac{ax}{7(x-y)}$

$\frac{x^{3}-y^{3}}{x+y}*\frac{x^{2}-y^{2}}{x^{2}+xy+y^{2}}=\frac{(x-y)(x^{2}+xy+y^{2})}{x+y}*\frac{(x+y)(x-y)}{x^{2}+xy+y^{2}}=(x-y)^{2}\\\\\frac{a^{2}-1 }{a^{3}+1}*\frac{a^{2}-a+1 }{a^{2}+2a+1 }=\frac{(a+1)(a-1)}{(a+1)(a^{2}-a+1)} *\frac{a^{2}-a+1 }{(a+1)^{2}}=\frac{a-1}{(a+1)^{2} }$

$\frac{x^{2}-10x+25}{3x+12}*\frac{x^{2}-16}{2x-10}=\frac{(x-5)^{2}}{3(x+4)}*\frac{(x+4)(x-4)}{2(x-5)}=\frac{(x-5)(x-4)}{6}=\frac{x^{2}-9x+20}{6}\\\\\frac{1-a^{2}}{4a+8b}*\frac{a^{2}+4ab+4b^{2}}{3-3a}=\frac{(1-a)(1+a)}{4(a+2b)}*\frac{(a+2b)^{2} }{3(1-a)}=\frac{(1+a)(a+2b)}{12}$

$\frac{y^{2}-25}{y^{2}+12y+36 }*\frac{3y+18}{2y+10}=\frac{(y-5)(y+5)}{(y+6)^{2} }*\frac{3(y+6)}{2(y+5)}=\frac{3y-15}{2y+12}\\\\\frac{b^{3}+8}{18b^{2}+27b}*\frac{2b+3}{b^{2}-2b+4}=\frac{(b+2)(b^{2}-2b+4)}{9b(2b+3)}*\frac{2b+3}{b^{2}-2b+4}=\frac{b+2}{9b}$