Question

x ori y unde:
x= √200 +2√72 + 4√32- 6√128
y= √108 + 2√12 - 4√75 + √300 - 2√192

Answer 1

 [tex]x=\sqrt{100*2}+2\sqrt{36*2}+4\sqrt16*2}-6\sqrt64*2}=\\ =(10+2*6+4*4-6*8)*\sqrt2=-10\sqrt2\;;\\ y=\sqrt{36*3}+2\sqrt{4*3}-4\sqrt{25*3}+\sqrt{100*3}-2\sqrt{64*3}=\\ =(6+2*2-4*5+10-2*8)*\sqrt3=-16\sqrt3\;;\\ x*y= +(10\sqrt2*16\sqrt3}=10*16*\sqrt{2*3}=160\sqrt6\;;[/tex]

Answer 2

$x= \sqrt{200} +2 \sqrt{72} +4 \sqrt{32} -6 \sqrt{128} \\ x=10 \sqrt{2} +2 \cdot 6 \sqrt{2} +4 \cdot 4 \sqrt{2} -6 \cdot 8 \sqrt{2} \\ x=10 \sqrt{2} +12 \sqrt{2} +16 \sqrt{2} -48 \sqrt{2} \\ x= -10 \sqrt{2} \\ y= \sqrt{108} +2 \sqrt{12} -4 \sqrt{75} + \sqrt{300} -2 \sqrt{192} \\ y=6 \sqrt{3} +2 \cdot 2 \sqrt{3} -4 \cdot 5 \sqrt{3} +10 \sqrt{3} -2 \cdot 8 \sqrt{3} \\ y=6 \sqrt{3} +4 \sqrt{3} -20 \sqrt{3} +10 \sqrt{3} -16 \sqrt{3} \\ y=-16 \sqrt{3}$
$x \cdot y=-10 \sqrt{2} \cdot (-16 \sqrt{3} )=160 \sqrt{6}$