Question

$1) frac{4+m}{4-m}*( frac{2m^2}{4+m}-m);2) (2+frac{a}{a+1})* frac{3a^2+3a}{12a+8};3)( frac{a}{a-5}-2a)/ frac{11-2a}{a-5}$

Answer 1

1) =$(2*m^2(4+m))/((4+m)*(4-m))-m*(4+m)/(4-m)=$$=2*m^2/(4-m) -m(4+m)/(4-m)=(m^2-4*m)/(4-m)=$

2)$=(6*a^2+6*a)/(12*a+8)+(3*a^3+3*a^2)/((a+1)(12*a+8))=$$=(9a^2+6*a)/(12*a+8)$

3)$=(a/(a-5)-2*a)/((a-5)/(11-2*a))=$$=(a-2*a^2+10*a)/(11-2*a)=a(11-2*a)/(11-2*a)=a$

Answer 2

Нетакой

Answer 3

падругому