If $x$  is real, then the expression $\frac{x^2+34x-71}{x^2+2x-7}$  can have no value between

The given expression is $\frac{x^2+34x-71}{x^2+2x-7}$

Let $\frac{x^2+34x-71}{x^2+2x-7}=m$ ,

$\Rightarrow x^2+34x-71-m\left(x^2+2x-7\right)=0$

$\Rightarrow \left(m-1\right)x^2+2\left(m-17\right)x+\left(71-7m\right)=0$

Since $x$  is real,   $\therefore$  Discriminant $\left(b^2-4ac\right)\ge 0$

$\Rightarrow$ $4{\left(m-17\right)}^2-4\left(m-1\right)\left(71-7m\right)\ge 0$

$\Rightarrow$  $m^2-14m+45\ge 0$  $\Rightarrow$  $\left(m-5\right)\left(m-9\right)\ge 0$

$\Rightarrow$ $m\le 5$  or $m\ge 9$

Hence $m$  can have no value between 5 or 9.