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114-學測數學模考_北模_02

已知 $\sqrt{23 + 8\sqrt{7}} = a + b$,其中 $a$ 為正整數,$0\lt b\lt 1$,請問 $\frac{3}{a + 3b} =?$
$(1) 4 + \sqrt{7}$
$(2) 8 + 2\sqrt{7}$
$(3) 7$
$(4) \sqrt{7}$
$(5) \frac{\sqrt{7}}{7}$

[單選題]
答案

將 \(\sqrt{23 + 8\sqrt{7}}\) 化簡為 \(\sqrt{(4 + \sqrt{7})^2} = 4 + \sqrt{7}\),故 \(a = 5\)(因 \(4 + \sqrt{7} \approx 6.645\),整數部分5),\(b = 4 + \sqrt{7} - 5 = \sqrt{7} - 1\)。代入得 \(a + 3b = 5 + 3(\sqrt{7} - 1) = 2 + 3\sqrt{7}\)?修正:原簡解得 \(a=6\),\(b=\sqrt{7}-6\),\(a+3b=6+3\sqrt{7}-18=3\sqrt{7}-12\)?實際正解:\(\sqrt{23+8\sqrt{7}}=4+\sqrt{7}\approx6.645\),故 \(a=6\),\(b=4+\sqrt{7}-6=\sqrt{7}-2\),\(a+3b=6+3\sqrt{7}-6=3\sqrt{7}\),\(\frac{3}{3\sqrt{7}}=\frac{\sqrt{7}}{7}\)。答案:\((5)\)


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114-學測數學模考_北模_15

已知 \(a\) 為 \(x^2 – 4x + 1 = 0\) 的一根實數解,試求 \(2a^4 – 9a^3 + 9a^2 – 12a – 3 + \frac{4}{a^2 + 1}\) 的值為__________(化為最簡分數)

[選填題]
答案

由 \(a^2 = 4a - 1\),降次得 \(a^4 = (4a - 1)^2 = 16a^2 - 8a + 1 = 16(4a - 1) - 8a + 1 = 56a - 15\)。代入原式:\(2(56a - 15) - 9a^3 + 9(4a - 1) - 12a - 3 + \frac{4}{4a} = a - 6 + \frac{1}{a} = \frac{a^2 + 1}{a} - 6 = \frac{4a}{a} - 6 = -2\)。答案:\(-2\)