Let $a, b, c$ be non-zero real numbers such that $a+b+c=0$. Let $q=a^2+b^2+c^2$ and $r=a^4+b^4+c^4$. Then,

For $x \in R$, the least value of $\frac{x^2-6x+5}{x^2+2x+1}$ is

If $x$ is real,the expression $\frac{x + 2}{2x^2 + 3x + 6}$ takes all values in the interval

If $\alpha, \beta$ are the roots of the equation $x^2 - px + q = 0$,then the quadratic equation whose roots are $(\alpha^2 - \beta^2)(\alpha^3 - \beta^3)$ and $\alpha^3\beta^2 + \alpha^2\beta^3$ is (where $S = p[p^4 - 5p^2q + 5q^2]$ and $P = p^2q^2(p^4 - 5p^2q + 4q^2)$).

Let $f(x)=(x-a)(x-b)-\left(\frac{a+b}{2}\right)$. If $f(x)=0$ has both non-negative roots,then the minimum value of $f(x)$ is:

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4+x^2+1=0$,then $\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=$