critical points are -3, -1,1, 3
Intervals (-\(\infty\), -3), (-3, -1), [-1,1], (1, 3), (3, \(\infty\))
The solution set is (-\(\infty\), -3), [-1,1], (3, \(\infty\))

\(\frac { 1 }{ 2+\sqrt { 3 } } \)=\(\frac { 1 }{ 2+\sqrt { 3 } } \times \frac { 2-\sqrt { 3 } }{ 2-\sqrt { 3 } } =\frac { 2-\sqrt { 3 } }{ 4-3 } =2-\sqrt { 3 } \)
\(\frac { 3 }{ 4-\sqrt { 5 } } =\frac { 3 }{ 4-\sqrt { 5 } } \times \frac { 4+\sqrt { 5 } }{ 4+\sqrt { 5 } } =\frac { 12+3\sqrt { 5 } }{ 16-5 } =\frac { 12+3\sqrt { 5 } }{ 11 } \)
\(\frac { 6 }{ 7-\sqrt { 8 } } =\frac { 6 }{ 7-\sqrt { 8 } } \times \frac { 7+\sqrt { 8 } }{ 7+\sqrt { 8 } } =\frac { 42+6\sqrt { 8 } }{ 49-8 } =\frac { 42+6\sqrt { 8 } }{ 41 } \)
∴ \(\frac { 1 }{ 2-\sqrt { 3 } } +\frac { 3 }{ 4-\sqrt { 5 } } +\frac { 6 }{ 7-\sqrt { 8 } } =2-\sqrt { 3 } +\frac { 12+3\sqrt { 5 } }{ 11 } +\frac { 42+6\sqrt { 8 } }{ 41 } \)

⇒  \({x+1\over x-1}>0\Rightarrow{{(x+1)(x-1)}\over{{(x-1)}^{2}}}>0\)
⇒ (x+1) (x-1)​> 0 ( \(\because\) (x-1)²>0 for all x≠1)​​​​​​

(x+1)(x-1) > 0
\(⇒ x \in (-∞,-1) \cup (1,∞)\)

Given quadratic equation is
5^2x-5^x+3 + 125 = 5x
\(\Rightarrow\) (5^x)²-5^x 5³- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-125 5^x- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-126.5^x+125=0
Let 5^x=y
\(\Rightarrow\)y²-126y + 125 = 0
\(\Rightarrow\)(y-1)(y-125) = 0
\(\Rightarrow\)    y = 1 or 125
\(\Rightarrow\)   5^x = 1 or 125
Case (i) When 5^x = 1 \(\Rightarrow\) 5⁰ \(\Rightarrow\) x = 0
Case (ii) When 5^x = 125 \(\Rightarrow\) 5^x = 5³\(\Rightarrow\) x = 3
\(\therefore\) The roots are 0, 3.

Given x > 2
Converting the into equation, we get x = 2. 
Clearly it is a line parallel to y-axis at a distance of 2 units from it.
Also, we find that the point (0, 0) does not satisfy the inequation x > 2
So, the region represented by the given equation is the shaded region.