If $ K + \frac{1}{K} - 2 = 0 $ and K > 0, then what is the value of $K^{17}+\frac{1}{k^{12}}$?
Answer & explanation
Correct answer: option 1
If $ K + \frac{1}{K} - 2 = 0 $
put k = 1
$K^{17}+\frac{1}{k^{12}}$ = 1 + 1 = 2
If $ K + \frac{1}{K} - 2 = 0 $ and K > 0, then what is the value of $K^{17}+\frac{1}{k^{12}}$?
Correct answer: option 1
If $ K + \frac{1}{K} - 2 = 0 $
put k = 1
$K^{17}+\frac{1}{k^{12}}$ = 1 + 1 = 2