Bài giải:

Ta có: $9{{\left( a+\frac{1}{36a} \right)}^{2}}-6.\left( a+\frac{1}{36a} \right)+1=0$

$\Rightarrow $${{\left[ 3\left( a+\frac{1}{36a} \right)-1 \right]}^{2}}$ = 0

$\Rightarrow $3. $\left( a+\frac{1}{36a} \right)$ = 1

$\Rightarrow $$a+\frac{1}{36a}$ = $\frac{1}{3}$

$\Rightarrow $a² - $\frac{1}{3}$a + $\frac{1}{36}$ = 0

$\Rightarrow $${{\left( a-\frac{1}{6} \right)}^{2}}=0$

$\Rightarrow $a = $\frac{1}{6}$

Vậy $\frac{1}{a}$= 6