If $x$ is a solution of the equation $\sqrt { 2 x + 1 } - \sqrt { 2 x - 1 } = 1 , \left( x \geq \frac { 1 } { 2 } \right)$, then $\sqrt { 4 x ^ { 2 } - 1 }$ is equal to :
(1) $\frac { 3 } { 4 }$
(2) $\frac { 1 } { 2 }$
(3) $2 \sqrt { 2 }$
(4) 2

We are to find the positive number $a$ satisfying $a ^ { 3 } = 9 + \sqrt { 80 }$.
Let us consider the positive number $b$ which satisfies $b ^ { 3 } = 9 - \sqrt { 80 }$. Then
$$\left\{ \begin{aligned} a ^ { 3 } + b ^ { 3 } & = \mathbf { A B } \\ a b & = \mathbf { C } \end{aligned} \right.$$
holds.
First, using (2), (1) can be transformed into
$$( a + b ) ^ { 3 } - \mathbf { D } ( a + b ) = \mathbf { A B } .$$
Then, setting $x = a + b$, we have
$$x ^ { 3 } - \mathrm { D } x = \mathrm { AB } .$$
Transforming this equation, we obtain
$$x ^ { 3 } - 27 = \mathbf { D } ( x - \mathbf { E } ) \text {, }$$
which gives
$$( x - \mathbf { F } ) \left( x ^ { 2 } + \mathbf { G } x + \mathbf{H} \right) = 0 .$$
From that we have $x = \mathbf{I}$ and hence
$$a + b = \mathbf{I} \text {. }$$
Thus, from (2), (3) and $a > b$, we have
$$a = \frac { \mathbf { J } + \sqrt { \mathbf { K } } } { \mathbf{L} } .$$

Given that
$$\frac { 1 } { \sqrt { x } - 6 } - \frac { 1 } { \sqrt { x } + 6 } = \frac { 3 } { 11 }$$
what is the value of $x$ ?
A $2 \sqrt { 15 }$ B $4 \sqrt { 5 }$ C $5 \sqrt { 2 }$ D $\sqrt { 58 }$ E 50 F 58 G 60 H 80

$$\sqrt { x } - \sqrt { y } = \sqrt { x + y - 1 }$$
Given this, what is the product $x \cdot y$?
A) $\frac { 1 } { 3 }$
B) $\frac { 1 } { 4 }$
C) $\frac { 3 } { 4 }$
D) $\frac { 2 } { 5 }$
E) $\frac { 4 } { 5 }$

$$\frac { 1 + \sqrt { a } } { 1 - a } - \frac { a } { 1 - \sqrt { a } } = \frac { 5 } { 3 }$$
Given this, what is a?
A) $\frac { 1 } { 2 }$
B) $\frac { 3 } { 2 }$
C) $\frac { 1 } { 4 }$
D) $\frac { 1 } { 9 }$
E) $\frac { 4 } { 9 }$

For a real number $\mathbf { x }$,
$$\sqrt { \frac { \sqrt { x } + 2 } { \sqrt { x } - 2 } } = \sqrt { x } + 2$$
Given that, what is x?
A) 3
B) 4
C) 5
D) 6
E) 7

Given that $x \neq - 1$ and
$$( x + 1 ) \sqrt { x - 1 } = \sqrt { 2 x + 2 }$$
What is $x$?
A) $\frac { 3 \sqrt { 3 } } { 2 }$
B) $\frac { 2 \sqrt { 3 } } { 3 }$
C) $\frac { 4 \sqrt { 2 } } { 3 }$
D) $\sqrt { 2 }$
E) $\sqrt { 3 }$

$$\frac { 1 } { \sqrt { 2 x } } + \frac { 4 } { \sqrt { 8 x } } = 6$$
Given this, what is $x$?
A) $\frac { 1 } { 2 }$
B) $\frac { 1 } { 4 }$
C) $\frac { 1 } { 6 }$
D) $\frac { 1 } { 8 }$
E) $\frac { 1 } { 12 }$

$\frac { 1 } { \sqrt { \mathrm{a} } } - \frac { 2 } { \sqrt { 9 \mathrm{a} } } = 1$\ Given this, what is a?\ A) $\frac { 1 } { 3 }$\ B) $\frac { 2 } { 3 }$\ C) $\frac { 1 } { 4 }$\ D) $\frac { 1 } { 9 }$\ E) $\frac { 4 } { 9 }$