Question

299 пожалуйста помогите мне

Answer 1

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

2) $\frac{a}{a^2-2a+1}- \frac{1}{1-a}* \frac{a}{a+1} - \frac{2}{a+1} = \frac{a}{(a-1)^2}+ \frac{1}{a-1}* \frac{a}{a+1} - \frac{2}{a+1} =$
$=\frac{a(a+1)}{(a-1)^2(a+1)}+ \frac{a(a-1)}{(a-1)^2(a+1)} - \frac{2(a-1)^2}{(a-1)^2(a+1)} =$
$=\frac{a^2+a+a^2-a-2(a^2-2a+1)}{(a-1)^2(a+1)}=\frac{2a^2-2a^2+4a-2}{(a-1)^2(a+1)}=\frac{4a-2}{(a-1)^2(a+1)}$

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

4) $( \frac{m}{n^2+mn} - \frac{m-n}{m^2+mn} ):( \frac{n^2}{m^3-mn^2} + \frac{1}{m+n} )=$
$=( \frac{m}{n(n+m)} - \frac{m-n}{m(m+n)} ):( \frac{n^2}{m(m^2-n^2)} + \frac{1}{m+n} )=$
$=( \frac{m^2}{mn(m+n)} - \frac{mn-n^2}{mn(m+n)} ):( \frac{n^2}{m(m-n)(m+n)} + \frac{m(m-n)}{m(m-n)(m+n)} )=$
$=\frac{m^2-mn+n^2}{mn(m+n)}:\frac{n^2+m^2-mn}{m(m-n)(m+n)}=\frac{m^2-mn+n^2}{mn(m+n)}* \frac{m(m-n)(m+n)}{n^2+m^2-mn}= \frac{m-n}{n}$