Question

Помогите пожалуйста у меня сор ​

Answer 1

№1.

$\displaystyle\\a)\frac{x^2+2x}{x^2}=\frac{x(x+2)}{x^2}=\frac{x+2}{x} \\\\b) \frac{y^2-7y}{y^2-49} =\frac{y(y-7)}{(y-7)(y+7)} =\frac{y}{y+7}$

№2

$\displaystyle a)\frac{3}{a} +\frac{a-3}{a+5} =\frac{3(a+5)+a(a-3)}{a(a+5)} =\frac{3a+15+a^2-3a}{a(a+5)} =\frac{a^2+15}{a^2+5a} \\\\b)\frac{3x^2}{x^2-9} -\frac{3x}{x+3} =\frac{3x^2-3x(x-3)}{(x-3)(x+3)} =\frac{3x^2-3x^2+9x}{(x-3)(x+3)} =\frac{9x}{x^2-9}$

№3

$\displaystyle\frac{y^2-12y+36}{y^2-36}\div \frac{10y-60}{y^2+6y} =\frac{(y-6)^2}{(y-6)(y+6)} \times\frac{y(y+6)}{10(y-6)} =\frac{y}{10} ;y=50\\\\50\div10=5$

№4

$\displaystyle\\\frac{x}{y} =d;\frac{y}{z}=\frac{1}{d} \\ \\x=dy;z=dy= > x=z= > \frac{x}{z}=1$