Question

18．若n為整數(shù)，且$\sqrt{{n}^{2}+9n+30}$是自然數(shù)，則n=-14或-7或-2或5．

Answer 1

分析 設(shè)$\sqrt{{n}^{2}+9n+30}$=p，再把等式兩邊同時(shí)乘以4，利用平方差公式把等式左邊化為兩個(gè)因式積的形式，列出關(guān)于p、n的方程組，求出n的值即可．

解答 解：∵設(shè)$\sqrt{{n}^{2}+9n+30}$=p（P為非負(fù)整數(shù)），則n²+9n+30=p²，
∴4n²+36n+120=4p²，
∴（2n+9）²+39=4p²，
∴（2p+2n+9）（2p-2n-9）=39，
∴$\left\{\begin{array}{l}2p+2n+9=1\\ 2p-2n-9=39\end{array}\right.$或$\left\{\begin{array}{l}2p+2n+9=39\\ 2p-2n-9=1\end{array}\right.$或$\left\{\begin{array}{l}2p+2n+9=3\\ 2p-2n-9=13\end{array}\right.$或$\left\{\begin{array}{l}2p+2n+9=13\\ 2p-2n-9=3\end{array}\right.$，
解得$\left\{\begin{array}{l}p=10\\ n=-14\end{array}\right.$或$\left\{\begin{array}{l}p=10\\ n=5\end{array}\right.$或$\left\{\begin{array}{l}p=4\\ n=-7\end{array}\right.$或$\left\{\begin{array}{l}p=4\\ n=-2\end{array}\right.$，
∴n=-14或-7或-2或5．
故答案為：-14或-7或-2或5．

點(diǎn)評(píng) 本題考查的是二次根式的性質(zhì)與化簡(jiǎn)，先根據(jù)題意把原式化為兩個(gè)因式積的形式是解答此題的關(guān)鍵．