ALGIRRAI - RICHARD - 02 Tople 6-1 Practico Select all the expressions that will result in a rational number. A $$ 5 \cdot \frac{\sqrt{3}}{2} $$ B. $$ 5 \frac{2}{3}+\frac{\sqrt{7}}{4} $$ C. $$ 5 \frac{2}{3}+\frac{17}{4} $$ D. $$ 5 \frac{2}{3}-\frac{\sqrt{16}}{2} $$ $$ E_{5 \frac{2}{3}+\frac{\sqrt{T}}{2}} $$

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1. For option A, we have
   $$
   5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}.
   $$
   Since $$\sqrt{3}$$ is irrational, multiplying it by 5 and dividing by 2 still leaves an irrational number.

2. In option B, note that the mixed number 
   $$
   5 \frac{2}{3} = 5 + \frac{2}{3}
   $$
   is rational, but the additional term 
   $$
   \frac{\sqrt{7}}{4}
   $$
   is irrational because $$\sqrt{7}$$ is irrational. The sum of a rational number and an irrational number is irrational.

3. Option C involves
   $$
   5 \frac{2}{3} + \frac{17}{4}.
   $$
   First, convert the mixed number to an improper fraction:
   $$
   5 \frac{2}{3} = 5 + \frac{2}{3} = \frac{15}{3} + \frac{2}{3} = \frac{17}{3}.
   $$
   Then, add the fractions:
   $$
   \frac{17}{3} + \frac{17}{4}.
   $$
   Find a common denominator (which is 12):
   $$
   \frac{17}{3} = \frac{17 \times 4}{12} = \frac{68}{12}, \quad \frac{17}{4} = \frac{17 \times 3}{12} = \frac{51}{12},
   $$
   so
   $$
   \frac{68}{12} + \frac{51}{12} = \frac{119}{12}.
   $$
   This is a rational number.

4. For option D, we have
   $$
   5 \frac{2}{3} - \frac{\sqrt{16}}{2}.
   $$
   Recognize that $$\sqrt{16} = 4$$, so
   $$
   \frac{\sqrt{16}}{2} = \frac{4}{2} = 2.
   $$
   With the mixed number converted as before:
   $$
   5 \frac{2}{3} = \frac{17}{3},
   $$
   subtracting gives:
   $$
   \frac{17}{3} - 2 = \frac{17}{3} - \frac{6}{3} = \frac{11}{3}.
   $$
   This result is rational.

5. Option E is
   $$
   5 \frac{2}{3} + \frac{\sqrt{T}}{2}.
   $$
   While $$5 \frac{2}{3}$$ is rational, the expression $$\frac{\sqrt{T}}{2}$$ depends on the value of $$T$$. It will be rational only if $$T$$ is such that $$\sqrt{T}$$ is a rational number (for example, if $$T$$ is a perfect square). Without additional information about $$T$$, we cannot guarantee that the result is rational.

Thus, the expressions that will always result in a rational number are options C and D.
Options C and D result in rational numbers.

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