Question

2010th root of (2√7-3√3) * 4020th root of (55+12√21)=?

Answer 1

Answer:

1

Step-by-step explanation:

$\sqrt[2010]{(2 \sqrt{7} - 3 \sqrt{3} )} \times \sqrt[4020]{55 + 12 \sqrt{21} } \\ \\ = \sqrt[2010]{(2 \sqrt{7} - 3 \sqrt{3} )} \times \sqrt[4020]{ 27 + 28+2 \times 2 \sqrt{3} \times 3 \sqrt{3} } \\ \\ = \sqrt[2010]{(2 \sqrt{7} - 3 \sqrt{3} )} \times \sqrt[4020]{(3 \sqrt{3}) {}^{2} + (2 \sqrt{7}) {}^{2} + 2\times 3 \sqrt{3} \times 2 \sqrt{3} } \\ \\ = \sqrt[2010]{(2 \sqrt{7} - 3 \sqrt{3} )} \times \sqrt[4020]{ (2\sqrt{7} + 3 \sqrt{3} ) {}^{2} } \\ \\ = \sqrt[2010]{(2 \sqrt{7} - 3 \sqrt{3} )} \times \sqrt[2010]{2 \sqrt{7} + 3 \sqrt{3} } \\ \\ = \sqrt[2010]{(2 \sqrt{7} - 3 \sqrt{3} )(2 \sqrt{ 7} + 3 \sqrt{3}) } \\ \\ \sqrt[2010]{( {2 \sqrt{7} )}^{2} - (3 \sqrt{3)} {}^{2} } \\ \\ = \sqrt[2010]{28 - 27} \\ \\ = \sqrt[2010]{1} \\ \\ = 1$

Answer 2

Answer:

1

Step-by-step explanation:

We can rewrite the equation as;

$(\sqrt[2010]{(2\sqrt{7}-3\sqrt{3}  })*(\sqrt[4020]{(55+12\sqrt{21} }))$

we apply BODMAS

and begin with operations within the brackets

The first bracket can be solved as;

$2\sqrt{7}-3\sqrt{3}=2(2.645751311)-3(1.732050808)\\  =0.095350199$

The first bracket can be solved as;

$55+12\sqrt{21}=55+12(4.582575695)\\ =109.9909083$

Now, the whole equation can be rewritten as;

$\sqrt[2010]{0.095350199} *\sqrt[4020]{109.9909083} \\=0.095350199^{\frac{1}{2010}}* 109.9909083^{\frac{1}{4020}\\ =0.99883143*1.001169937\\=1$