Question

_______________________________РЕШИТЬ______________

Answer 1

$a) frac{a+b}{a-2b} * frac{2b-a}{(a-b)(a+b)}= -frac{a-2b}{(a-2b)(a-b)}=- frac{1}{a-b}$

$b) frac{5x-y}{6y-5x}* frac{-(6y-5x)}{-(5x-y)}= frac{-1}{-1}=1$

$frac{(4t-7)^3}{(4t-2)^4} * frac{(4t-2)^4}{(4t-7) ^{4} }= frac{1}{4t-7}$

$- frac{3 k^{2}*p^5 }{8f*l^2*r^3}* frac{8f^2*l*r}{3k^2*p^3}=- frac{p^2*f}{l*r ^{2} }$

$frac{(i+1)^2}{(i-1)^3}* frac{(i-1)^2}{(i+1)^3}= frac{1}{(i+1)(i-1)}= frac{1}{i^2-1}$

$frac{4l-5k}{j(g+7)}* frac{-j^2*3(g+7)}{2z(4l-5k)}=- frac{3j}{2z}$

Answer 2

а
$(a+b)/(a-2b)*(a-2b)/([(b-a)(b+a)=1/(b-a)$
б
$(5x-y)/(6y-5x)*(6y-5x)/(5x-y)=1$
в
$(4t-7)^3/(4t-2)^4*(4t-2)^4/(4t-7)^4=1/(4t-7)$
г
$-3k^2p^5/8fl^2r^3 * 8f^2lr/3k^2p^3=-fp^2/lr^2$
д
$(i+1)^2/(i-3)^3*(i-2)^2/(i+1)^3=1/[(i-1)(i+1)]=1/(i^2-1)$
е
-(4l-5k)/[j(g+7)]*3j²(3g+7)/[2z(4l-5k)]=-3j/2z