Since an even power of any real number cannot be negative, the even root of a negative number is not a real number.

For \( n \in Z^+\), in order for \(\large \sqrt[2n]{a} \) to exist, a ≥ 0 must hold.

Example:

If \( x^4 =-16\), then \( x \notin R\). Because the fourth power of any real number x cannot be -16. \( \sqrt[4]{ -16} \notin R, \; \sqrt{ -7} \notin R\)

However,

If \(x^3 =-8 \), then \( x = \sqrt[3]{-8 } \in R\).

Question 1

If \[ \large A= \frac{\sqrt[3]{x} + \sqrt[4]{x-3}}{1+\sqrt{5-x}} \]

how many integer values can x take for A to be a real number?

\[ \text{A)} 1 \quad \text{B) } 2 \quad \text{C) } 3 \quad \text{D) } 4 \quad \text{E)} 5 \]

Solution:

Since the degrees of the roots \[ \sqrt[4]{x-3 } \; \text{and} \; \sqrt{ 5-x} \] are even numbers,

\( x-3 ≥ 0 \) and \( 5-x ≥ 0 \) must hold.

\[ x-3 ≥ 0 \Rightarrow x≥0 \]

\[ 5-x≥ 0 \Rightarrow 5≥x \]

Thus, \( 3≤ x ≤ 5 \). Based on this, the integer values x can take are 3, 4, and 5, which means there are 3 values.

\(\textbf{Answer: C} \)

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