1. **State the problem:** We need to find a rational expression for the total amount of food per fasting person, given: - Rice per person: $\frac{2}{x}$ kilograms - Dates per person: $\frac{3}{x+2} + 23$ kilograms where $x$ is the number of volunteers. 2. **Write the total amount of food per person:** $$\text{Total} = \frac{2}{x} + \left(\frac{3}{x+2} + 23\right)$$ 3. **Combine the terms:** $$\text{Total} = \frac{2}{x} + \frac{3}{x+2} + 23$$ 4. **Find a common denominator for the fractional parts:** The denominators are $x$ and $x+2$, so the common denominator is $x(x+2)$. Rewrite each fraction: $$\frac{2}{x} = \frac{2(x+2)}{x(x+2)} = \frac{2x + 4}{x(x+2)}$$ $$\frac{3}{x+2} = \frac{3x}{x(x+2)}$$ 5. **Add the fractions:** $$\frac{2x + 4}{x(x+2)} + \frac{3x}{x(x+2)} = \frac{2x + 4 + 3x}{x(x+2)} = \frac{5x + 4}{x(x+2)}$$ 6. **Add the constant 23:** Since 23 is not a fraction, to combine it with the fraction, write 23 as $\frac{23x(x+2)}{x(x+2)}$: $$23 = \frac{23x(x+2)}{x(x+2)}$$ 7. **Total expression:** $$\text{Total} = \frac{5x + 4}{x(x+2)} + \frac{23x(x+2)}{x(x+2)} = \frac{5x + 4 + 23x(x+2)}{x(x+2)}$$ 8. **Simplify numerator:** $$23x(x+2) = 23x^2 + 46x$$ So numerator is: $$5x + 4 + 23x^2 + 46x = 23x^2 + (5x + 46x) + 4 = 23x^2 + 51x + 4$$ 9. **Final rational expression:** $$\frac{23x^2 + 51x + 4}{x(x+2)}$$ 10. **Compare with given options:** None of the options match this expression exactly, but the problem likely expects the sum of the fractional parts only (rice + dates without the constant 23), which is: $$\frac{5x + 4}{x(x+2)}$$ This matches option 4: $\frac{5x + 4}{x(x+2)}$ **Answer:** $\boxed{\frac{5x + 4}{x(x+2)}}$