\sqrt{11+4 \sqrt{6} } = a \sqrt{2} +b \sqrt{3} nilai b-a =........
Takamori37
1 votes
Thanks 1
Takamori37
Lupa hehe,
Udah diedit kok
rasiyah
blom gtu ??
rasiyah
y ws udh mkasi y?????????
Takamori37
Oke
rasiyah
bntu aq lgi donk
Takamori37
Post aja
rasiyah
\sqrt{19+8 \sqrt{3} } : \sqrt{3} +1 dapat d sederhanakan menjadi.........
rasiyah
udah py aq tmbh lgy dsni ws
Takamori37
\sqrt{19+8\sqrt{3}} = \sqrt{19+2\sqrt{48}}
Didapat \sqrt{16}+\sqrt{3}=4+\sqrt{3}
(4+\sqrt{3})/(\sqrt{3}+1) = ((3+sqrt(3))(sqrt(3)-1)/2
1/2 (3sqrt(3)-3+3-sqrt(3)) = 1/2 sqrt(3) = sqrt(3)
rasiyah
bingung aqnya,,,coba anda jwb aja yg sudh aq pos