Autorzy:Adam Makowski, Tomasz Masłowski, Anna Toruńska
Wydawnictwo:WSiP
Rok wydania:2016
Wykonaj działania. a) (√6+5√3)∙√34.67 gwiazdek na podstawie 6 opinii

`a) \ \ (sqrt6+5sqrt3)*sqrt3=sqrt6*sqrt3+5sqrt3*sqrt3=sqrt18+5sqrt9=sqrt(9*2)+5*3=sqrt9*sqrt2+15=3sqrt2+15` 

`b) \ \ (5sqrt8-2sqrt2)*sqrt18=(5sqrt(4*2)-2sqrt2)*sqrt(9*2)=(5sqrt4*sqrt2-2sqrt2)*sqrt9*sqrt2=(5*2*sqrt2-2sqrt2)*3sqrt2=` 

`=(10sqrt2-2sqrt2)*3sqrt2=10sqrt2*3sqrt2-2sqrt2*3sqrt2=30sqrt4-6sqrt4=30*2-6*2=60-12=48` 

`c) \ \ (3sqrt12+sqrt15)*2sqrt3=3sqrt12*2sqrt3+sqrt15*2sqrt3=6sqrt36+2sqrt45=6*6+2sqrt(9*5)=36+2*sqrt9*sqrt5= ` 

`=36+2*3*sqrt5=36+6sqrt5` 

`d) \ \ (2root(3)3+root(3)6)*root(3)9=2root(3)3*root(3)9+root(3)6*root(3)9=2root(3)27+root(3)54=2*3+root(3)(2*27)=6+root(3)2*root(3)27=6+root(3)2*3=6+3root(3)2` 

`e) \ \ (5root(3)4+root(3)12)*2root(3)2=5root(3)4*2root(3)2+root(3)12*2root(3)2=10root(3)8+2root(3)24=10*2+2root(3)(8*3)=20+2*root(3)8*root(3)3=20+4root(3)3` 

`f) \ \ (root(3)25+2root(3)50)*root(3)5=root(3)25*root(3)5+2root(3)50*root(3)5=root(3)125+2root(3)250=5+2root(3)(125*2)=5+2root(3)125*root(3)2=5+2*5*root(3)2=5+10root(3)2`