Check the correct answer:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}=$

This simple example illustrates the order of operations, which states that multiplication and division take precedence over addition and subtraction, and that operations within parentheses come first,

Let's say we have a fraction and a whole number (every whole number) between which a division operation takes place, meaning - we can relate to the fraction and the whole number as fractions in their simplest form, through which a division operation occurs, thus we can write the given fraction in the following form:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\
\downarrow\\
\big((2^2-3)^{15}+4^2\big):(15+2)-(3^2-2^2):5$We emphasize this by stating **that we should relate to the fractions that are in the numerator and those in the denominator separately**, as if they exist in their simplest form,

**Let's return to the original fraction in question**, meaning - in its given form, and simplify it, simplifying separately the different fractions that are in the numerator and those in the denominator (if simplification is needed), this is done in accordance with the order of operations mentioned above and in a systematic way,

We start with the first numerator from the left in the given fraction, noting that in this case it **changes the fraction in the denominators that are in multiplication**, therefore, we start with this fraction, **this in accordance with the aforementioned order of operations**, noting further that in this fraction in the denominators (which are in multiplication of 15) there exists a multiplication, therefore, we start calculating its numerical value in multiplication and then perform the subtraction operation that is in the denominators:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\
\frac{(4-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\
\frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5} \\$** We continue** with the fraction we received in the previous step and simplify the numerators and the denominators in the fraction, **this is done in accordance with the order of operations mentioned above**, therefore, we start calculating their numerical values in multiplication and then perform the division and subtraction operations that are in the numerators and in the denominators:

$\frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\
\frac{1+16}{17}-\frac{9-4}{5}= \\
\frac{17}{17}-\frac{5}{5}\\$** We continue **and simplify the fraction we received in the previous step, **again, in accordance with the order of operations mentioned above**, therefore, we perform the division operation of the denominators, **this is done systematically**, and then perform the subtraction operation:

$\frac{17}{17}-\frac{5}{5}=\\
\frac{\not{17}}{\not{17}}-\frac{\not{5}}{\not{5}}=\\
1-1=\\
0$

** We conclude **with this, the steps of simplifying the given fraction, we received that:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\
\frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5} =\\
\frac{1+16}{17}-\frac{9-4}{5}= \\
0$__Therefore, the correct answer is answer D.__