8−(2+2×3)21:49+2=
\( \frac{21:\sqrt{49}+2}{8-(2+2\times3)}= \)
\( (\frac{1}{4})^2+\frac{1}{16}= \)
\( (\frac{1}{2})^2+(\frac{1}{3})^2+\frac{1}{4}= \)
\( (\frac{1}{4})^2+\frac{1}{16}= \)
In the numerator we solve the square root exercise:
In the denominator we solve the exercise within parentheses:
The exercise we now have is:
We solve the exercise in the numerator of fractions from left to right:
We obtain the exercise:
Since it is impossible for the denominator of the fraction to be 0, it is impossible to solve the exercise.
Cannot be solved
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