$\text{The possible value of}x\text{satisfying}\frac{143÷13\times x-3\times 3}{6^2-7\times 5+x^2}=1\text{is:}$

The correct answer is 1.

We are given the equation:
$\frac{143÷13\times x-3\times 3}{6^2-7\times 5+x^2}=1$
To find the possible value of $x$, we can simplify the numerator and denominator step by step.

Step 1: Simplify the Numerator
Following the order of operations (BODMAS/PEMDAS):
First, perform the division:
$143÷13=11$
Next, perform the multiplications:
$11\times x=11x$
and
$3\times 3=9$
Thus, the simplified numerator is:
$11x-9$

Step 2: Simplify the Denominator
Evaluate the exponent and multiplication terms:
$6^2=36$
and
$7\times 5=35$
Substituting these values back into the denominator expression gives:
$36-35+x^2=1+x^2$

Step 3: Solve the Equation
Substitute the simplified numerator and denominator back into the original fraction:
$\frac{11x-9}{1+x^2}=1$
Cross-multiply to remove the fraction:
$11x-9=1+x^2$
Rearrange all terms to one side to form a standard quadratic equation:
$x^2-11x+10=0$
Factor the quadratic equation:
$(x-1)(x-10)=0$
This yields two solutions:
$x=1$
or
$x=10$

Among the given options, the possible value of $x$ present is 1.