Question

CРОЧНО!! ^5√√17-7 ∙ ^5√√17+7

Answer 1

1)

$\sqrt[5]{\sqrt{17-7} } \times\sqrt[5]{\sqrt{17-7}} =\sqrt[5]{\sqrt{17-7} \times\sqrt{17-7} } =\sqrt[5]{17-7} =\sqrt[5]{10}$

2)

$\sqrt[5]{\sqrt{17} -7}\times \sqrt[5]{\sqrt{17} -7} =\sqrt[5]{(\sqrt{17} -7)(\sqrt{17}-7 )} =\\\\=\sqrt[5]{(\sqrt{17} -7)^2}=\sqrt[5]{17-14\sqrt17+49} } =\sqrt[5]{66-14\sqrt{17} }$

Answer 2

To simplify the expression ^5√√17-7 ∙ ^5√√17+7, let's first break it down step by step.

1. Start with ^5√√17. This means you're taking the fifth root of the fourth root of 17. So, √√17 is the fourth root of 17, and ^5√ means you're taking the fifth root of that.

   ^5√√17 = ^5√(17^(1/4))

2. Now, let's simplify the expression ^5√(17^(1/4)) - 7 ∙ ^5√(17^(1/4)) + 7.

3. We can see that ^5√(17^(1/4)) is a common factor, so we can factor it out:

   ^5√(17^(1/4)) * (1 - 7) + 7

4. Next, simplify the numbers within the parentheses:

   ^5√(17^(1/4)) * (-6) + 7

5. Finally, you can multiply -6 by ^5√(17^(1/4)) and then add 7:

   -6 * ^5√(17^(1/4)) + 7

So, the simplified expression is: -6 * ^5√(17^(1/4)) + 7.

Answer 3

To calculate ^5√√17 - 7 ∙ ^5√√17 + 7, you can break it down into smaller steps:

1. Calculate ^5√√17:

   • ^5√√17 is the fifth root of the fourth root of 17.
   • The fourth root of 17 (√√17) is the same as ^4√17.
   • So, you need to find ^5√(^4√17).

2. Calculate ^5√(^4√17):

   • To do this, you can use the properties of radicals. The fifth root of the fourth root is the same as the ^20√17.
   • So, ^5√(^4√17) = ^20√17.

Now that you have ^5√(^4√17) = ^20√17, you can substitute this back into the original expression:

^20√17 - 7 ∙ ^20√17 + 7

Now, you can calculate this expression:

^20√17 - 7 * ^20√17 + 7 = (^20√17 - 7 * ^20√17) + 7

Factor out ^20√17:

^20√17 * (1 - 7) + 7

Now, calculate 1 - 7:

^20√17 * (-6) + 7

Finally, multiply -6 by ^20√17 and add 7:

-6 * ^20√17 + 7

This is your simplified expression. If you need a numerical approximation, you would need to calculate the value of ^20√17 using a calculator or a computer program.