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\(f(x)=\sqrt{3 + \sqrt{5 + \sqrt{x}}}\)

\(x\ge0\)

Smallest value of the function would be \(\sqrt{3 + \sqrt{5}}\)

Maximum value is infinity.

\(a + b\sqrt{c} = \sqrt{3 + \sqrt{5}}\)

\(a + b\sqrt{5} = \sqrt{3 + \sqrt{5}}\)

(a + bsqrt(5))^2 = 3 + sqrt(5)

a^2 + 5b^2 + 2absqrt(5) = 3 + sqrt(5)

a^2 + 5b^2 = 3

sqrt(5) = 2absqrt(5)

2ab = 1

ab = 1/2

You can do it from here... Domain: [a + bsqrt(c), inf)

Get a and b, or just use the sqrt(3 + sqrt(5))

First we want to get y to cancel out.

Equation 1 can be rewritten as 12x - 18y = -6

Equation 2 can be rewritten as 2kx - 18y = 6

Subtracting equation 2 from equation 1:

(12 - 2k)x = -12

(6-k)x = -6

If k = 6, then 0 = -6 which is false.

Simplify:  
(2x + 6x - 9) + (8 - 4x - 2x)
2x + 6x - 9 + 8 - 4x - 2x
8x - 6x - 1
2x - 1

no clue

         a   b   c

1  = (6, 0, 0)
2  = (4, 1, 0)
3  = (2, 2, 0)
4  = (0, 3, 0)
5  = (3, 0, 1)
6  = (1, 1, 1)
7  = (0, 0, 2)

a(0)==1

a(1) == -1

a(2)== - 1 + 2*1 ==1

a(3)==1 + 2*- 1 ==-1

a(4)== - 1 + 2*1 == 1

a(5)==1 + 2*- 1 == - 1........and so on

Then the sequence is: 1,  - 1,  1,  - 1,  1,  - 1.........etc.

All EVEN terms end in 1 and all ODD terms end in - 1

So: a(100)== - 1 + 2*1 ==1

For the following values of n, the expression will have integral values:

n ==1,  2,  13,  26

[1, 1, 1, 1, 1, 2, 2, 2] ==1 combination

8! / [5!*3!] ==56 permutations.

C(n) is a positive integer, but there is no positive integer that, when cubed, equals 16.

Like so:

Angle ETF = 72 degrees.

There are 35 ways.

cos angle ABM = 1/4.

They intersect in the first and fourth quadrants.

Angle ETF = 60 degrees.