Question

Aratati ca numerele a,b,c sunt intregi :
a= 4√10- 2√8- (6+ 4√10)
b= 2(4+√21)- 6√7 + 2√7 * (3- √3)
c= 4√3(√8- √3) - 2√2 * (3√3-√2)

Efectuati:
a) 2√132 : √11 + 14√30 : (2√10) + 9√21 : √7 - 6 * √3
b) (3√2 - √5) * √3 + √2 * (√3 - √20) + √5 * (√3 + √8)
c) √48 * (2√75 - √27 + 2√27) - √48 * √12
d) √243 * √147 - 2√3 * 7√3 + 3√8 (5√72 + 6√128)

Answer 1

$1).a=4 \sqrt{10} -2 \sqrt{8} -(6+4 \sqrt{10} ) \\ a=4 \sqrt{10} -2*2 \sqrt{2} -6-4 \sqrt{10} \\ a=4 \sqrt{10} -4 \sqrt{2} -6-4 \sqrt{10} \\ a=-4 \sqrt{2} -6 \\ b=2(4+ \sqrt{21} )-6 \sqrt{7} +2 \sqrt{7} (3- \sqrt{3} ) \\ b=8+2 \sqrt{21} -6 \sqrt{7} +6 \sqrt{7} -2 \sqrt{21} \\ b=8$
$c=4 \sqrt{3} ( \sqrt{8} - \sqrt{3} )-2 \sqrt{2} (3 \sqrt{3} - \sqrt{2} ) \\ c=4 \sqrt{3} (2 \sqrt{2} - \sqrt{3} )-6 \sqrt{6} +4 \\ c=8 \sqrt{6} -12-6 \sqrt{6} +4 \\ c=2 \sqrt{6} -8$
$2a).2 \sqrt{132} : \sqrt{11} +14 \sqrt{30} :2 \sqrt{10} +9 \sqrt{21} : \sqrt{7} -6 \sqrt{3} = \\ =2 \sqrt{12} +7 \sqrt{3} +9 \sqrt{3} -6 \sqrt{3} =2*2 \sqrt{3} +7 \sqrt{3} +9 \sqrt{3} -6 \sqrt{3} = \\ =4 \sqrt{3} +7 \sqrt{3} +9 \sqrt{3} -6 \sqrt{3} =14 \sqrt{3}$
$b).(3 \sqrt{2} - \sqrt{5} )* \sqrt{3} + \sqrt{2} ( \sqrt{3} - \sqrt{20} )+ \sqrt{5} ( \sqrt{3} + \sqrt{8} )= \\ =3 \sqrt{6} - \sqrt{15} + \sqrt{6} - \sqrt{40} + \sqrt{15} + \sqrt{40} =4 \sqrt{6}$
$c). \sqrt{48} (2 \sqrt{75} - \sqrt{27} +2 \sqrt{27} )- \sqrt{48} * \sqrt{12}= \\ =2 \sqrt{3600} - \sqrt{1296} +2 \sqrt{1296} - \sqrt{576} = \\ =2*60-36+2*36-24=120-36+72-24=132$
$d). \sqrt{243}* \sqrt{147} -2 \sqrt{3} *7 \sqrt{3} +3 \sqrt{8} (5 \sqrt{72} +6 \sqrt{128} )= \\ = \sqrt{35721} -42+15 \sqrt{576} +18 \sqrt{1024} = \\ =189-42+15*24+18*32=189-42+360+576=1083$