If $x$ is real,then the difference between the greatest and least values of $\frac{x^2-x+1}{x^2+x+1}$ is

The least value of $\frac{x^2y^2 - 2x^2y + 2x^2 + 2xy - 2x + 1}{x^2y + x}$ is $\lambda$,where $x, y \in R^+$ and $x^2y + x \neq 0$. Then:

If $x = \frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right)$, then the value of $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$ is equal to

For $0 < c < b < a$,let $(a+b-2c)x^2 + (b+c-2a)x + (c+a-2b) = 0$ and $\alpha \neq 1$ be one of its roots. Then,among the two statements:
$(I)$ If $\alpha \in (-1, 0)$,then $b$ cannot be the geometric mean of $a$ and $c$.
$(II)$ If $\alpha \in (0, 1)$,then $b$ may be the geometric mean of $a$ and $c$.

Let $\alpha, \beta$ be the roots of the equation $x^2 - 3x + r = 0$, and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2 + 3x + r = 0$. If the roots of the equation $x^2 + 6x = m$ are $2\alpha + \beta + 2r$ and $\alpha - 2\beta - \frac{r}{2}$, then $m$ is equal to:

If the equation $x^8 - kx^2 + 3 = 0$ has a real solution,then the least integral value of $k$ is-