What will come at place of x, (x < 10) for $$\frac{{\left( {132 \div 12 \times x - 3 \times 3} \right)}}{{\left( {{5^2} - 6 \times 4 + {x^2}} \right)}} = 1?$$

$$\eqalign{ & \frac{{132 \div 12 \times x - 3 \times 3}}{{{5^2} - 6 \times 4 + {x^2}}} = 1 \cr & 11 \times x - 9 = 25 - 24 + {x^2} \cr & 11x - 9 = 1 + {x^2} \cr & {x^2} + 1 + 9 - 11x = 0 \cr & {x^2} - 11x + 10 = 0 \cr & {x^2} - 10x - x + 10 = 0 \cr & x\left( {x - 10} \right) - 1\left( {x - 10} \right) = 0 \cr & \left( {x - 10} \right)\left( {x - 1} \right) = 0 \cr & x = 10,\,1 \cr & {\text{For the condition}}\left( {x < 10} \right) \cr & x = 1 \cr} $$