What will be the possible value of $x$ that satisfies the given equation?
$$\frac{(143÷13\times x-3\times 3)}{(6^2-7\times 5+x^2)}=1$$

Let's simplify the numerator and denominator of the given equation step-by-step:
Step 1: Simplify the Numerator

$$\text{Numerator}=143÷13\times x-3\times 3$$

According to BODMAS rules, perform division first:

$$143÷13=11$$

Now substitute it back:

$$\text{Numerator}=11\times x-9=11x-9$$

Step 2: Simplify the Denominator

$$\text{Denominator}=6^2-7\times 5+x^2$$

$$=36-35+x^2$$

$$=1+x^2$$

Step 3: Solve the Equation

$$\frac{11x-9}{1+x^2}=1$$

Cross-multiplying gives:

$$11x-9=1+x^2$$

Rearranging the terms into a standard quadratic equation format ($a{x}^2+bx+c=0$):

$$x^2-11x+1+9=0$$

$$x^2-11x+10=0$$

Step 4: Factorize the Quadratic Equation

$$x^2-10x-x+10=0$$

$$x(x-10)-1(x-10)=0$$

$$(x-1)(x-10)=0$$

Thus, the possible values of $x$ are $x=1$ or $x=10$.
From the given options, $1$ is available.