If x is a prime number and (a + b + c - 1 + x)^2 + |1/(1 - x)

\(\sqrt{(a + b + c - 1 + x)^2 }+ |\frac{1}{(1 - x)} - a - b - c|+\sqrt{c^2} = 0\)­

Looking at the equation one notices that: \(\sqrt{(a + b + c - 1 + x)^2 }\), \(|\frac{1}{(1 - x)} - a - b - c|\) and \(\sqrt{c^2}\) will never result in a negative.­ The only way adding three of these together to get \(0\) is if each equals \(0\).

The easiest to recognise is that \(c = 0\). Plugging that into the above: 

\(\sqrt{(a + b - 1 + x)^2 } + |\frac{1}{(1 - x)} - a - b|= 0\)­

Now as we know that both of the parts equal zero, we can rewrite it as: \(a + b - 1 + x + \frac{1}{(1 - x)} - a - b= 0\)­. This is as the square or the squareroot of \(0\) is always \(0\) and as \(0\) is neither positive nor negative one does require the absolute signs.

\(a + b - 1 + x + \frac{1}{(1 - x)} - a - b= 0\)­

\( x + \frac{1}{(1 - x)} = 1\)­

\(\frac{x-x^2+1}{1-x} = 1\)

\(x-x^2+1 = 1 - x\)

\(x^2 - 2x = 0\)

\(x(x-2)=0\)

Therefore \(x = 0\) or \(x = 2\). As we are told that \(x\) is prime, \(x = 0\) and therefore \(x+c = 2\)

ANSWER D

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