Question

Scoateti factorii de sub radical,apoi calculati:
a) √18+√288) - (√98+√200)
b) 4√18+5√108-2√72
c)3√162-5√12-4√75+2√50
d) -4√216+3√96-√54+2√600

Answer 1

$a).( \sqrt{18} + \sqrt{288} )-( \sqrt{98} + \sqrt{200} )= (3 \sqrt{2} +12 \sqrt{2} )-(7 \sqrt{2} +10 \sqrt{2} )= \\ =15 \sqrt{2} -17 \sqrt{2} =-2 \sqrt{2} \\ b).4 \sqrt{18} +5 \sqrt{108} -2 \sqrt{72} =4*3 \sqrt{2} +5*6 \sqrt{2} -2*6 \sqrt{2} = \\ =12 \sqrt{2} +30 \sqrt{2} -12 \sqrt{2} =30 \sqrt{2}$
$c).3 \sqrt{162} -5 \sqrt{12} -4 \sqrt{75} +2 \sqrt{50} = \\ =3*9 \sqrt{2} -5*2 \sqrt{3} -4*5 \sqrt{3} +2*5 \sqrt{2} = \\ =27 \sqrt{2} -10 \sqrt{3} -20 \sqrt{3} +10 \sqrt{2} =37 \sqrt{2} -10 \sqrt{3} \\ d).-4 \sqrt{216} +3 \sqrt{96} - \sqrt{54} +2 \sqrt{600} = \\ =-4*6 \sqrt{6} +3*4 \sqrt{6} -3 \sqrt{6} +2*10 \sqrt{6} = \\ =-24 \sqrt{6} +12 \sqrt{6} -3 \sqrt{6} +20 \sqrt{6} =5 \sqrt{6}$