第127页

信息发布者：
D
C
$-\frac{1}{a}$
2
$m^{2}+3n^{2}$
$2mn$
(2)解：$\because a + 4\sqrt{3}=(m + n\sqrt{3})^{2}，$
$\therefore a + 4\sqrt{3}=m^{2}+3n^{2}+2mn\sqrt{3}。$
$\therefore a=m^{2}+3n^{2}，$$2mn = 4，$于是$mn = 2。$
$\because m，$$n$均为正整数，$\therefore m = 1，$$n = 2$或$m = 2，$$n = 1。$
$\therefore a=1^{2}+3\times2^{2}=13$或$a=2^{2}+3\times1^{2}=7。$
$\therefore a$的值为13或7。
(3)解：原式$=\sqrt{5 + 2\sqrt{5}+1}=\sqrt{(\sqrt{5})^{2}+2\times\sqrt{5}\times1 + 1^{2}}$
$=\sqrt{(\sqrt{5}+1)^{2}}=\sqrt{5}+1。$
解：由$\sqrt{a - 2025}$有意义可知$a - 2025\geq0，$从而$a\geq2025。$
$\because\sqrt{(2024 - a)^{2}}+\sqrt{a - 2025}=a，$
$\therefore a - 2024+\sqrt{a - 2025}=a。$
$\therefore\sqrt{a - 2025}=2024。$$\therefore a - 2025=2024^{2}。$
$\therefore a=2025 + 2024^{2}。$
$\therefore\frac{a - 1}{2024}=\frac{2025 + 2024^{2}-1}{2024}=\frac{2024 + 2024^{2}}{2024}=2025。$
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