2 times root sign 3 times (1 part root sign 6 + 3 part root sign 2)

2 times root sign 3 times (1 part root sign 6 + 3 part root sign 2)

2 times root sign 3 times (1 part root sign 6 + 3 part root sign 2)
=2√3×（√6+√2/3）
=6√2+2√6/3;
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(5 under 3-radical) / 2 Comparison 1 / 2

3 / 2 - (heel 5 /) 2
5 > 4, so follow 5 > 2
(below 5) / 2 is greater than 1
So it's less than 1 / 2

Calculation: [1 / (2 + root 3)] + under root sign (4 + 2 root sign 3)

The original formula = (2 - √ 3) / (2 + √ 3) (2 - √ 3) + √ (√ 3 + 1) square
=(2-√3)/(4-3)+√3+1
=2-√3+√3+1
=3

Why a + B ≥ 2 root sign ab What I don't understand the most is why the AB under the root sign is twice, instead of the AB under the root sign twice

The condition is a > 0, b > 0
The square is greater than or equal to 0
So a > 0, b > 0
Then (√ a - √ b) 2 ≥ 0
a-2√ab+b≥0
a+b≥2√ab

What is the derivative of (x ^ 2 + 1) under the radical?

Root sign (x square + 1) x
X
---------
√(x^2+1)

How to find the integral of 2 of 1 + X under radical

If x = TANU, then DX = sec? UDU, then the original formula = ∫ sec? UDU = ∫ secud (TANU) = secutanu - ∫ tanud (secu) = secutanu - ∫ tan? Usecudu = secutanu - ∫ sec? U-1) secudu = secutanu - ∫ sec? UDU +

Reduction 2 / (5 root sign 2 + root 3)

2 / (5 root number 2 + root number 3)
=2 (5 root number 2-root number 3) / (5 root number 2 + root number 3) (5 root number 2-root number 3)
=2（5√2-√3）/(50-3)
=2（5√2-√3）/47

Simplification: 1 / [(2 radical 1) + (root 2)] + 1 / [(3 radical 2) + (2 radical 3)] + 1 / [(4 radical 3) + (3 radical 4)] + +1 / [(100 root number 99) + (99 root number Simplification: 1 / [(2 radical 1) + (root 2)] + 1 / [(3 radical 2) + (2 radical 3)] + 1 / [(4 radical 3) + (3 radical 4)] + +1 / [(100 root number 99) + (99 root number 100)] should be very detailed

1/[(n+1)√n+n√(n+1)]
=1/√n-1/√(n+1)
1/(2√1+√2)+1/(3√2+2√3)+...+1/(100√99+99√100)
=1-1/√2+1/√2-1/√3+1/√3-1/√4+...+1/√98-1/√99+1/√99-1/√100
=1-1/√100
=1-1/10
=9/10

Calculation or simplification: 2 / 1 + 2 + 2 / 2 + 3 + 4 ^ 2 / 99 + 100

2 / 1 + 2 + 2 / 2 / 2 + 3 + 4 ^ 2 / 99 + 100
=2 (root 2-1) / (1 + 2 under root) (2-1) + 2 (3-2) / (2 + 3) (3-2) + 2 (4-3) / (3-4-3) ^ 2 (100-99) / (99 + 100) (100-99)
=2 [root 2-1 + root 3-root 2 + root 4-root 3 +. + root 100 root 99]
=2 × (root 100-1)
=2×9
=18

: (sin20) ^ 2 + (cos80) ^ 2 + root 3cos80cos20 =?

(sin20) ^ 2 + (cos80) ^ 2 + radical 3cos80cos20 = 2 (sin20) ^ 2 + √ 3cos80cos20 = 0.5 * √ 3sin40 + 1-cos40