Which of the following can be the value of k, if $\frac{88÷8\times k-3\times 3}{6^2-7\times 5+k^2}=1$?

The correct option is 1, 10.

To find the possible values of $k$, we start with the given equation:

$\frac{88÷8\times k-3\times 3}{6^2-7\times 5+k^2}=1$

Let's simplify the numerator and the denominator separately using the order of operations (BODMAS/PEMDAS).

Step 1: Simplify the Numerator
The numerator is $88÷8\times k-3\times 3$.
First, perform the division:
$88÷8=11$
Now substitute this back and perform the multiplications:
$11\times k=11k$
$3\times 3=9$
So, the simplified numerator is:
$11k-9$

Step 2: Simplify the Denominator
The denominator is $6^2-7\times 5+k^2$.
First, calculate the exponent:
$6^2=36$
Next, perform the multiplication:
$7\times 5=35$
Now substitute these values back into the expression:
$36-35+k^2$
Combine the constant terms:
$1+k^2$

Step 3: Solve the Equation
Substitute the simplified numerator and denominator back into the main equation:
$\frac{11k-9}{1+k^2}=1$

Multiply both sides by $1+k^2$ to eliminate the fraction:
$11k-9=1+k^2$

Rearrange the terms to form a standard quadratic equation ($a{k}^2+bk+c=0$):
$k^2-11k+1+9=0$
$k^2-11k+10=0$

Step 4: Factor the Quadratic Equation
We need to find two numbers that multiply to $10$ and add up to $-11$. These numbers are $-1$ and $-10$.
Rewrite the quadratic equation by factoring:
$(k-1)(k-10)=0$

Set each factor to zero to find the roots:
$k-1=0\Rightarrow k=1$
$k-10=0\Rightarrow k=10$

Thus, the possible values of $k$ are 1 and 10.