Question

Stie cineva?     ......              

Answer 1

$\sqrt{(a- \sqrt{96}) ^{2} } + \sqrt{(b+ \sqrt{216}) ^{2} }+ \sqrt{(c- \sqrt{24}) ^{2} } \leq 0$

$\sqrt{(a- \sqrt{96}) ^{2} }= (a- \sqrt{96}) ^{2* \frac{1}{2} } =a- \sqrt{96}$
$\sqrt{(b+ \sqrt{216}) ^{2} }= (b+ \sqrt{216}) ^{2* \frac{1}{2} } =b+ \sqrt{216}$
$\sqrt{(c- \sqrt{24}) ^{2} }= (c-\sqrt{24}) ^{2* \frac{1}{2} } =c- \sqrt{24}$

$(a- \sqrt{96}) + (b+ \sqrt{216}) + (c- \sqrt{24}) \leq 0$

$\sqrt{96} = \sqrt{ 2^{5}*3 } = 2^{2} \sqrt{2*3} =4 \sqrt{6}$
$\sqrt{216} = \sqrt{ 2^{3} * 3^{3} } =2*3* \sqrt{2*3} =6 \sqrt{6}$
$\sqrt{24} = \sqrt{2 ^{3}*3 } =2 \sqrt{2*3} =2 \sqrt{6}$

$a- 4\sqrt{6} + b+6 \sqrt{6} + c-2 \sqrt{6} \leq 0$

$a+b+c- 6\sqrt{6} +6 \sqrt{6} \leq 0$

$a+b+c \leq 0$