ABC26GN0819 · Simplification

Let $\frac{128 \div 16 \times x-7 \times 2}{7^{2}-8 \times 6+x^{2}}=1$. Then, $8 x-7 \times 2=49-48+x^{2} \Leftrightarrow 8 x-14=1+x^{2} \Leftrightarrow x^{2}-8 x+15=0$ $$\begin{aligned} & \Leftrightarrow x^{2}-3 x-5 x+15=0 \Leftrightarrow x(x-3)-5(x-3)=0 \\ & \Leftrightarrow(x-3)(x-5)=0 \Leftrightarrow x=3 \text { or } x=5 . \end{aligned}$$ Hence, missing number is 3 or 5.