Powers and Roots: Using fractions

$\frac{21:\sqrt{49}+2}{8-(2+2\times3)}=$

In the numerator we solve the square root exercise:

$\sqrt{49}=7$

In the denominator we solve the exercise within parentheses:

$(2+2\times3)=$

$2+6=8$

The exercise we now have is:

$\frac{21:7+2}{8-8}=$

We solve the exercise in the numerator of fractions from left to right:

$21:7=3$

$3+2=5$

We obtain the exercise:

$\frac{5}{8-8}=\frac{5}{0}$

Since it is impossible for the denominator of the fraction to be 0, it is impossible to solve the exercise.

Cannot be solved

$(\frac{1}{4})^2+\frac{1}{16}=$

1/8

$(\frac{1}{2})^2+(\frac{1}{3})^2+\frac{1}{4}=$

11/18

$(\frac{1}{4})^2+\frac{1}{16}=$

1/8