Since, $a$ and $b$ are roots of $x^2-7x-1=0$, so

$a^2-7a-1=0$

$\Rightarrow a^2-1=7a$

$\Rightarrow a^4+1-2a^2=49a^2$

$\Rightarrow a^4+1=51a^2 ...\left(1\right)$

Similarly,

$b^4+1=51b^2 ...\left(2\right)$

Now,

$\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$

$=\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}}$

$=\frac{51a^{19}+51b^{19}}{a^{19}+b^{19}}$

$=51$