Question

(Radical )
√3(√12-2√27)

√48-2√3(2-5√12)

(3+√3)(2+√3)

(√6+√5) (totul la puterea 2) -√120

merci anticipat

Answer 1

$a). \sqrt{3} ( \sqrt{12} -2 \sqrt{27} )= \sqrt{3} (2 \sqrt{3} -2 \cdot 3 \sqrt{3} )= \sqrt{3} (2 \sqrt{3} -6 \sqrt{3} )= \\ =2 \cdot 3 - 6 \cdot 3=6-18=-12 \\ b). \sqrt{48} -2 \sqrt{3} (2-5 \sqrt{12} )=4 \sqrt{3} -2 \sqrt{3} (2-5 \cdot 2 \sqrt{3} )= \\ =4 \sqrt{3} -2 \sqrt{3} (2-10 \sqrt{3} )=4 \sqrt{3} -4 \sqrt{3} +20 \cdot 3=60 \\ c).(3+ \sqrt{3} )(2+ \sqrt{3} )=6+3 \sqrt{3} +2 \sqrt{3} +3=9+5 \sqrt{3}$
$d).( \sqrt{6} + \sqrt{5} )^2- \sqrt{120} = \sqrt{6} ^2+2 \cdot \sqrt{6} \cdot \sqrt{5} + \sqrt{5}^2-2 \sqrt{30} = \\ =6+2 \sqrt{30} +5-2 \sqrt{30} =11$