Which of the given numbers are negative?
(a) $10-3 \sqrt{11}$
(c) $18-5 \sqrt{13}$
(e) $10 \sqrt{26}-51$
(b) $3 \sqrt{11}-10$
(d) $51-10 \sqrt{26}$
We have $3 \sqrt{11}=\sqrt{9 \times 11}=\sqrt{99}$. Since $100>99$, we get $10=\sqrt{100}>\sqrt{99}=3 \sqrt{11}$, therefore, $10-3 \sqrt{11}>0$ and $3 \sqrt{11}-102600$, we obtain

$$
51=\sqrt{2601}>\sqrt{2600}=10 \sqrt{26}
$$

thus, $51-10 \sqrt{26}>0$ and $10 \sqrt{26}-51<0$. Finally, $18^{2}=324<325=25 \times 13$ ensures that $18<5 \sqrt{13}$, so $18-5 \sqrt{13}<0$.

The negative numbers are (b) $3 \sqrt{11}-10$, (c) $10 \sqrt{26}-51$ and (e) $18-5 \sqrt{13}$.