Question

Ayuda con estos problemas

Answer 1

La primera:

$\frac{y^2+5y+6}{y-1}\cdot \frac{y^2-1}{\left(x+3\right)^2}\\\\=\frac{\left(y^2+5y+6\right)\left(y^2-1\right)}{\left(y-1\right)\left(x+3\right)^2}\\\\=\frac{\left(y+2\right)\left(y+3\right)\left(y+1\right)\left(y-1\right)}{\left(x+3\right)^2\left(y-1\right)}\\\\=\frac{\left(y+2\right)\left(y+3\right)\left(y+1\right)}{\left(x+3\right)^2}$

La segunda:

$\frac{2m^2-3m-4}{6m+3}\cdot \frac{3m+6}{m^2-4}\\\\=\frac{3}{m-2}\cdot \frac{2m^2-3m-4}{6m+3}\\\\=\frac{3\left(2m^2-3m-4\right)}{3\left(m-2\right)\left(2m+1\right)}\\\\=\frac{2m^2-3m-4}{\left(2m+1\right)\left(m-2\right)}$

La tercera:

$\frac{x^2-xy+y-x}{x^2+2x+1}\cdot \frac{3}{6x^2-9x+3}\\\\=\frac{x^2-xy-x+y}{x^2+2x+1}\cdot \frac{1}{2x^2-3x+1}\\\\=\frac{\left(x^2-xy+y-x\right)\cdot \:1}{\left(x^2+2x+1\right)\left(2x^2-3x+1\right)}\\\\=\frac{x^2-xy+y-x}{\left(x^2+2x+1\right)\left(2x^2-3x+1\right)}\\\\=\frac{\left(x-1\right)\left(x-y\right)}{\left(x+1\right)^2\left(x-1\right)\left(2x-1\right)}\\\\=\frac{x-y}{\left(x+1\right)^2\left(2x-1\right)}$

La cuarta (ultima):

$\frac{x^2+7x+10}{x^2-6x-7}\cdot \frac{x^2-3x-4}{x^2+2x-15}\\\\=\frac{\left(x^2+7x+10\right)\left(x^2-3x-4\right)}{\left(x^2-6x-7\right)\left(x^2+2x-15\right)}\\\\=\frac{\left(x+2\right)\left(x+5\right)\left(x+1\right)\left(x-4\right)}{\left(x^2-6x-7\right)\left(x^2+2x-15\right)}\\\\=\frac{\left(x+2\right)\left(x+5\right)\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x-7\right)\left(x-3\right)\left(x+5\right)}\\\\=\frac{\left(x+2\right)\left(x-4\right)}{\left(x-7\right)\left(x-3\right)}$