Question

решите во ложениях там обведено

Answer 1

$(frac{4m}{2-m}-m):(2+frac{m^2+4}{m-2})=(frac{4m}{2-m}-frac{m(2-m)}{2-m}):(frac{2(m-2)}{m-2}+frac{m^2+4}{m-2})=$

$=(frac{4m-2m+m^2}{2-m}):(frac{2m-4+m^2+4}{m-2})=$

$=(frac{m^2+2m}{2-m}):(frac{m^2+2m}{m-2})=$

$=(frac{m^2+2m}{2-m})cdot(frac{m-2}{m^2+2m})=frac{m-2}{2-m}=-1$

 

$frac{x^3y-xy^3}{xy^2-x2y}=frac{xy(x^2-y^2)}{xy(y-1)}=frac{(x-y)(x+y)}{y-1}$

$frac{(1/3-1/4)(1/3+1/4)}{1/4-1}=frac{(4/12-3/12)(4/12+3/12)}{1/4-4/4}=frac{1/12cdot7/12}{-3/4}=frac{7}{-18}=-frac{7}{18}$

Answer 2

$(frac{4m}{2-m}-m):(2+frac{m^2+4}{m-2})=frac{4m-2m+m^2}{2-m}*frac{m-2}{2m-4+m^2+4}=$

$=-frac{2m+m^2}{m-2}*frac{m-2}{2m+m^2}=-1$

 

№3 не понятно, там пример через x и y , а подставлять a и b