1. **Problem 221:** Find the value(s) of $x$ such that $\frac{1}{x} = \frac{1}{x+1} = \frac{1}{x+4}$. 2. **Step 1:** Since all three fractions are equal, set $\frac{1}{x} = \frac{1}{x+1}$. For two fractions $\frac{1}{a} = \frac{1}{b}$ to be equal, either $a = b$ or both denominators are undefined. 3. **Step 2:** Equate denominators: $x = x + 1$ which is impossible, so no solution here. 4. **Step 3:** Similarly, set $\frac{1}{x} = \frac{1}{x+4}$, which implies $x = x + 4$, also impossible. 5. **Step 4:** Since all three fractions are equal, the only way is if the denominators are such that the fractions are undefined or equal to zero. But $\frac{1}{x}$ is undefined at $x=0$, $\frac{1}{x+1}$ undefined at $x=-1$, and $\frac{1}{x+4}$ undefined at $x=-4$. 6. **Step 5:** Check if any of the given options make all three fractions equal: - At $x=0$, $\frac{1}{0}$ undefined, $\frac{1}{1}=1$, $\frac{1}{4}=0.25$ not equal. - At $x=-1$, $\frac{1}{-1}=-1$, $\frac{1}{0}$ undefined, no. - At $x=-2$, $\frac{1}{-2}=-0.5$, $\frac{1}{-1}=-1$, $\frac{1}{2}=0.5$ no. - At $x=-3$, $\frac{1}{-3}=-0.333$, $\frac{1}{-2}=-0.5$, $\frac{1}{1}=1$ no. - At $x=-4$, $\frac{1}{-4}=-0.25$, $\frac{1}{-3}=-0.333$, $\frac{1}{0}$ undefined no. 7. **Step 6:** Since none of the options satisfy the equality, the only possibility is that the fractions are all equal to zero, which is impossible for $\frac{1}{x}$ type fractions. So no solution from the options. --- 8. **Problem 222:** Simplify $\left(\frac{1}{2}\right)^{-3} \left(\frac{1}{4}\right)^{-2} \left(\frac{1}{16}\right)^{-1}$. 9. **Step 1:** Recall the rule $a^{-n} = \frac{1}{a^n}$ and $\left(\frac{1}{a}\right)^{-n} = a^n$. 10. **Step 2:** Rewrite each term: - $\left(\frac{1}{2}\right)^{-3} = 2^3 = 8$ - $\left(\frac{1}{4}\right)^{-2} = 4^2 = 16$ - $\left(\frac{1}{16}\right)^{-1} = 16^1 = 16$ 11. **Step 3:** Multiply the results: $8 \times 16 \times 16 = 8 \times 256 = 2048$. 12. **Step 4:** Express $2048$ as a power of $\frac{1}{2}$: Since $2^{11} = 2048$, then $2048 = 2^{11} = \left(\frac{1}{2}\right)^{-11}$. 13. **Final answer:** $\boxed{\left(\frac{1}{2}\right)^{-11}}$.