1. **Problem statement:** Given positive numbers $x, y, z, t$ satisfying the equality $$x^2 = y^3 = \sqrt{z} = \frac{1}{t} > 1,$$ determine the correct ordering of $t, y, x, z$. 2. **Step 1: Express all variables in terms of a common variable.** Let $$k = x^2 = y^3 = \sqrt{z} = \frac{1}{t} > 1.$$ Then: - $$x = \sqrt{k} = k^{\frac{1}{2}}$$ - $$y = k^{\frac{1}{3}}$$ - $$z = (\sqrt{z})^2 = k^2$$ - $$t = \frac{1}{k}$$ 3. **Step 2: Analyze the order of $t, y, x, z$.** Since $k > 1$: - $$t = \frac{1}{k} < 1$$ - $$y = k^{\frac{1}{3}} > 1$$ - $$x = k^{\frac{1}{2}} > 1$$ - $$z = k^2 > 1$$ 4. **Step 3: Compare $t$ with others.** Since $t = \frac{1}{k} < 1$ and all others are greater than 1, we have: $$t < y, x, z$$ 5. **Step 4: Compare $y, x, z$.** Since $k > 1$, the function $k^a$ increases with $a$: - $$y = k^{\frac{1}{3}}$$ - $$x = k^{\frac{1}{2}}$$ - $$z = k^2$$ Because $$\frac{1}{3} < \frac{1}{2} < 2,$$ it follows that: $$y < x < z$$ 6. **Step 5: Combine all inequalities:** $$t < y < x < z$$ **Final answer:** (A) $t < y < x < z$