Question

помогите пожалуйста по алгебре,плачу не мало!​

Answer 1

$\frac{a^2+3a}{9-a^2}=\frac{a(a+3)}{(3-a)(3+a)}=\frac{a}{3-a}\\\\\\\frac{m^5+m^3}{m^3}=\frac{m^3(m^2+1)}{m^3}=m^2+1=(2\sqrt3)^2+1=4\cdot 3+1=13\\\\\\\frac{7n}{n^2-9n}=\frac{7n}{n(n-9)}=\frac{7}{n-9}\\\\\\\frac{x^2-10x+25}{25-x^2}=\frac{(x-5)^2}{(5-x)(5+x)}=\frac{(x-5)^2}{-(x-5)(x+5)}=-\frac{x-5}{x+5}=\frac{5-x}{x+5}$