Question

помогите пожалуйста1​

Answer 1

№1

а)$\frac{y^2-y}{y^2}=\frac{y(y-1)}{y^2}=\frac{y-1}{y}$

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

№2

a)

$\frac{4}{a} +\frac{a-4}{a+3}=\frac{4(a+3)+a(a-4)}{a(a+3)}=\frac{4a+12+a^2-4a}{a(a+3)}=\frac{a^2+12}{a(a+3)}$

b)

$\frac{2x^2}{x^2-4}-\frac{2x}{x+2}= \frac{2x^2}{(x-2)(x+2)}-\frac{2x}{x+2}=\frac{2x^2-2x(x-2)}{(x-2)(x+2)} =\frac{2x^2-2x^2+4x}{(x-2)(x+2)} =\frac{4x}{x^2-4}$

№3

$\frac{y^2-8y+16}{y^2-16}:\frac{10y-40}{y^2+4y}=\frac{(y-4)^2}{(y-4)(y+4)}* \frac{y(y+4)}{10(y-4)} =\frac{(y-4)y(y+4)}{(y+4)10(y-4)} =\frac{y}{10}$

При у = 60

$\frac{y}{10} =\frac{60}{10} =6$

№4

$\frac{x}{y} *\frac{y}{z} =\frac{x*y}{y*z}=\frac{x}{z} \\B*\frac{1}{B} =\frac{B*1}{B} =1\\\\\frac{x}{z}=1$