If x = 7 then ,what is the value of
$$\frac{x^\frac{1}{2} + x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1-\sqrt{x}}=?$$

$$\frac{x^\frac{1}{2} + x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1-\sqrt{x}}$$

$$\frac{\left(\sqrt{\ x}+\frac{1}{\sqrt{\ x}}\right)}{1-x}+\frac{\left(1-\frac{1}{\sqrt{\ x}}\right)}{1-\sqrt{\ x}}$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{\left(x+1\right)}{1-x}+\left(\frac{1}{\sqrt{\ x}}\cdot\frac{\left(\sqrt{\ x}-1\right)}{1-\sqrt{\ x}}\right)$$

$$\left(\frac{1}{\sqrt{\ x}}\cdot\frac{x+1}{1-x}\right)-\frac{1}{\sqrt[\ ]{x}}$$

$$\frac{1}{\sqrt{\ x}}\cdot\left(\frac{\left(x+1\right)}{1-x}-1\right)$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{\left(x+1-1+x\right)}{1-x}$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{2x}{1-x}$$

$$2\ \frac{\sqrt{\ x}}{1-x}$$

Substitute 7 in place of x.

$$2\ \frac{\sqrt{\ 7}}{1-7}$$

That will be $$-\frac{\sqrt{\ 7}}{3}$$