Упражнение 129

\({\largeа)}\ \frac{x^2-10x+25}{3x+12}\cdot\frac{x^2-16}{2x-10}=\frac{ (x-5)^2\cdot(x-4)(x+4)}{ 3(x+4)\cdot2(x-5)}=\frac{ (x-5)(x-4)}{6}=\frac{x^2-9x+20}{6};\)

\({\largeб)}\ \frac{1-a^2}{4a+8b}\cdot\frac{a^2+4ab+4b^2}{3-3a}=\frac{ (1-a)(1+a)\cdot(a+2b)^2}{ 4(a+2b)\cdot3(1-a)}=\frac{ (1+a)(a+2b)}{12}=\frac{a^2+2ab+a+2b}{12};\)

\({\largeв)}\ \frac{y^2-25}{y^2+12y+36}\cdot\frac{3y+18}{2y+10}=\frac{ (y-5)(y+5)\cdot3(y+6)}{ (y+6)^2\cdot2(y+5)}=\frac{ 3(y-5)}{ 2(y+6)}=\frac{3y-15}{2y+12};\)

\({\largeг)}\ \frac{b^3+8}{18b^2+27b}\cdot\frac{2b+3}{b^2-2b+4}=\frac{ (b+2)(b^2-2b+4)\cdot(2b+3)}{ 9b(2b+3)\cdot(b^2-2b+4)}=\frac{b+2}{9b}.\)