MaxWong

Observation: What's the minimal polynomial of \(x\)?

\(2x = 1 + \sqrt 3\\ 2x - 1 = \sqrt 3\\ (2x - 1)^2 = 3\\ 4x^2 - 4x - 2 = 0\\ 2x^2 - 2x - 1 = 0\)

Having that in mind:

\(\quad 4x^3 + 2x^2 - 8x + 7\\ = 4x^3 - 4x^2 - 2x + 6x^2 - 6x + 7\\ =2x(2x^2 - 2x - 1) + 3(2x^2 - 2x - 1) + 10\\ = 2x(0) + 3(0) + 10\\ = \boxed{10}\)

Let \(t = x - \dfrac1x\).

\(t^2 + 2 = x^2 + \dfrac1{x^2} = 171\)

\(t^2 = 169\)

\(t = \pm 13\)

\(x - \dfrac1x = \pm13\)

Therefore there are two possible values.

There is something missing in the question.

"equals () of sum of four numbers"

There must be something replacing the brackets

Each interior angle of a regular octagon is 135^o.

By Law of Cosines,

\(AC^2 = 8^2 + 8^2 - 2(8)(8)\cos135^\circ\\ AC^2 = 64\cdot \left(2 + \sqrt2\right)\\ AC = 8\sqrt{2+\sqrt 2}\)

\(\left(m + \dfrac1m\right)^2 = 9^2 \\ m^2 + 2m\cdot \dfrac1m + \left(\dfrac1m\right)^2 = 81\\ m^2 + \dfrac1{m^2} = 81-2 = \boxed{79}\)

Let \(x \equiv 11^{-1}\pmod{1000}\)

Now, because \(11^{\varphi(1000)} \equiv 1 \pmod{1000}\), where \(\varphi(x)\) is the Euler totient function,

\(x\equiv 11^{\varphi(1000) - 1} \pmod{1000}\)

We proceed to find the value of \(\varphi(1000)\).

\(\varphi(1000) = 1000\left(1 - \dfrac12\right)\left(1 - \dfrac15\right) = 400\)

Therefore

\(x\equiv 11^{399} \pmod{1000}\)

\(x\equiv 1331^{133} \pmod{1000}\\ x\equiv 331^{133}\pmod{1000}\)

Now, 

\(331^{133}\equiv 3 \pmod 8 \)

\(331^{133}\equiv 81^{33}\equiv3^{32} = 7^6 \cdot 9 \equiv 9216 \equiv 91\pmod{125}\)

By CRT, \(x \equiv 331^{133}\equiv 91 \pmod{1000}\)

This means \(11^{-1} \equiv 91 \pmod{1000}\).

Let f(x) = ax² + bx + c.

\(\begin{cases}a + b + c = 2\\4a + 2b + c = -7\\16a + 4b + c = 13\end{cases}\)

Solving by matrix method(or whatever method you like):

\(\begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1&1&1\\4&2&1\\16&4&1\end{bmatrix}^{-1}\cdot \begin{bmatrix}2\\-7\\13\end{bmatrix} = \begin{bmatrix}\dfrac{19}3\\-28\\\dfrac{71}3\end{bmatrix} \)

Therefore \(f(x) = ax^2 + bx + c = \dfrac{19}3 x^2 - 28x + \dfrac{71}3\)

Substituting x = 5 gives the answer. I believe you can do it from here.

For the general form of straight line(ax + by + c = 0), the slope is equal to -a/b.

Now substitute a = 5, b = 2.

Considering the general term of the expansion:

\(\left(\sqrt x + \dfrac5x\right)^9 = \displaystyle\sum_{k = 0}^9 \binom{9}k\cdot \left(\sqrt x\right)^k \left(\dfrac5x\right)^{9-k}\)

Simplifying,

\(\left(\sqrt x + \dfrac5x\right)^9 = \displaystyle\sum_{k = 0}^9 \binom{9}k \cdot 5^{9-k}\cdot x^{3k/2-9}\)

When the term is constant, power of x is 0.

\(\dfrac{3k}2 - 9 = 0\\ k = 6\)

Now, substitute k = 6 into the general term, you get the constant term.

(Note that if k was not an integer, the constant term would be 0.)