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leya454
 3 years ago
olve this radical expression (show work)
http://i1052.photobucket.com/albums/s441/leya454/mathhel_zps1ebfb197.png
leya454
 3 years ago
olve this radical expression (show work) http://i1052.photobucket.com/albums/s441/leya454/mathhel_zps1ebfb197.png

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