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leya454
Group Title
olve this radical expression (show work)
http://i1052.photobucket.com/albums/s441/leya454/mathhel_zps1ebfb197.png
 one year ago
 one year ago
leya454 Group Title
olve this radical expression (show work) http://i1052.photobucket.com/albums/s441/leya454/mathhel_zps1ebfb197.png
 one year ago
 one year ago

This Question is Closed

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
There is no solving here, you can't solve an expression, only simplify/reduce one...$$\frac{k+1}{k^2+6k+9}+\frac4{k+3}\frac{6}{k1}=\frac{(k+1)(k1)}{(k+3)^2(k1)}+\frac{4(k+3)(k1)}{(k+3)^2(k1)}\frac{6(k+3)^2}{(k+3)^2(k1)}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{(k+1)(k1)+4(k+3)(k1)+6(k+3)^2}{(k+3)^2(k1)}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{k^21+4k^2+8k12+6k^2+36k+54}{(k+3)^2(k1)}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{11k^2+44k41}{k^3+5k^2+3k9}$$
 one year ago
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