Question

$\sqrt[2010]{20102\sqrt{7}-3\sqrt{3}}\times \sqrt[4020]{402055+12\sqrt{21}}=?$

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

Answer: (B)

$1$

$\sqrt[2010]{20102\sqrt{7}-3\sqrt{3}}\times \sqrt[4020]{402055+12\sqrt{21}}$
$=\sqrt[2010]{20102\sqrt{7}-3\sqrt{3}}\times \sqrt[4020]{4020{\left(2\sqrt{7}\right)}^2+{\left(3\sqrt{3}\right)}^2+2\times 2\sqrt{7}\times 3\sqrt{3}}$
$=\sqrt[2010]{20102\sqrt{7}-3\sqrt{3}}\times \sqrt[4020]{4020{(2\sqrt{7}+3\sqrt{3})}^2}$
$=\sqrt[2010]{20102\sqrt{7}-3\sqrt{3}}\times \sqrt[2010]{2010(2\sqrt{7}+3\sqrt{3})}$
$=\sqrt[2010]{2010(2\sqrt{7}-3\sqrt{3})(2\sqrt{7}+3\sqrt{3})}$
$=\sqrt[2010]{2010({\left(2\sqrt{7}\right)}^2-{\left(3\sqrt{3}\right)}^2)}$
$=\sqrt[2010]{2010(28-27)}$
$=\sqrt[2010]{20101}$
$=1$.

Therefore, option $B$ is correct.