109年全國高中職教甄聯招-數學科詳解

教育部受託辦理109學年度
公立高級中等學校教師甄選

數學科試題

第壹部分：選擇題
一、單選題

解：

$$\cases{S_n的最大值為S_7 \\ |a_7| < |a_8|} \Rightarrow \cases{a_7> 0\\ a_8 < 0 \\ a_7+a_8 < 0} \Rightarrow \cases{2a_7= a_6+a_8= a_5+a_9 = \cdots=a_1+a_{13} > 0\\ 2a_8= a_7+a_9= a_6+a_{10} = \cdots=a_1+a_{15} < 0 \\ a_7+a_8 = a_6+a_9 = \cdots = a_1+a_{14} < 0} \\ \Rightarrow \cases{S_{13} >0 \\ S_{15}< 0 \\ S_{14} < 0} \Rightarrow n=13，故選\bbox[red, 2pt]{(B)}$$

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2. 已知某種快篩試劑對某病毒的檢驗，其「偽陰率」為8%（即帶原者做檢驗有8%的機會呈陰性反應，其他呈陽性反應），而「偽陽率」為1% （即未帶原者做檢驗有1%的機會呈陽性反應，其他呈陰性反應）。某地區經快篩試劑篩檢後呈現陽性反應的民眾中有 2% 為此病毒的帶原者，則此地區病毒的帶原者占全部人口的比例約為何？

(A) 2% (B)0.2% (C) 0.02% (D) 0.002% 。

解：

$$\cases{帶原比例p\\ 未帶原比例1-p} \Rightarrow {帶原且檢測為陽性 \over 檢測為陽性} =0.02 \Rightarrow { p\times 0.92 \over p\times 0.92 + (1-p)\times 0.01} =0.02 \\ \Rightarrow {0.92p \over 0.91p+0.01} =0.02 \Rightarrow {4509 \over 5000}p= {1\over 5000} \Rightarrow p={1\over 4509} \approx 0.00022 = 0.022\%，故選\bbox[red, 2pt]{(C)}$$

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解：
$$\cases{f(x)=10-x \\ a=0 \\ b=10} \Rightarrow \cases{A=10\times 10\div 2 = 50\\ R_1=(b-a)f(a) = 10\times 10=100 } \Rightarrow R_1 > A，故選\bbox[red,2pt]{(C)}$$

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解：

$$\cases{中圓O_2(0,25),半徑r_2=25\\ 大圓O_1(100,100),半徑r_1=100 \\ 小圓O_3(x,r_3),半徑r_3} \Rightarrow \overline{AB} =\overline{AO_3} + \overline{O_3B}=r_1=100 \\ \Rightarrow \sqrt{\overline{O_2O_3}^2 -\overline{O_2A}^2} + \sqrt{\overline{O_1O_3}^2 -\overline{O_1B}^2} =100 \\\Rightarrow \sqrt{(r_3+25)^2-(25-r_3)^2} +\sqrt{(100+r_3)^2-(100-r_3)^2} =100 \\ \Rightarrow 10\sqrt{r_3} +20\sqrt{r_3}=100 \Rightarrow \sqrt r_3={10\over 3} \Rightarrow r_3={100\over 9}，故選\bbox[red,2pt]{(A)}$$

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解：
$$\cases{O_1(1,2k),r_1=\sqrt 2\\ O_2(k,-1),r_2=2\sqrt 2} \Rightarrow \overline{O_1O_2} > r_1+r_2 \Rightarrow (k-1)^2+(2k+1)^2 > 3\sqrt 2 \\ \Rightarrow (5k-8)(k+2) > 0 \Rightarrow k>8/5 或k < -2，故選\bbox[red,2pt]{(D)}$$

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解： $$h(f(x))=g(x+1) \Rightarrow h(3x+1)=3(x+1)^2+3 = 3x^2+6x+6 = {1\over 3}(3x+1)^2 +{4\over 3}(3x+1)+{13\over 3} \\ \Rightarrow h(x)={1\over 3}x^2 +{4\over 3}x+{13\over 3} \Rightarrow h'(x)={2\over 3}x+{4\over 3}  \Rightarrow h'(0)={4\over 3} =\lim_{x\to 0}{h(x)-h(0) \over x}，故選 \bbox[red, 2pt]{(C)}$$

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解：

$$斜線區域面積S= \int_0^{2a} (x-{x\over a}(x-a)) \;dx =\int_0^{2a} -{1\over a}x^2+2x\;dx = \left. \left[ -{1\over 3a}x^3+x^2 \right] \right|_0^{2a} \\ =-{8\over 3}a^2+4a^2 = {4\over 3}a^2  \Rightarrow 兩圖形所圍面積=2S= {8\over 3}a^2，故選\bbox[red,2pt]{(A)}$$

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解：

$$延長\overline{EF}及延長\overline{DC},相交於G點(見上圖),則\cases{ \angle A = \angle EDG=90^\circ \\ \angle AFE = \angle EGD \\ \overline{AE} =\overline{ED}} \Rightarrow \triangle AEF \cong \triangle DEG \\ \Rightarrow 令\overline{AF}=a = \overline{GD} \Rightarrow \overline{EF} =\sqrt{a^2+{1\over 4}} = \overline{GE}; 由於\triangle GCF為等腰 \Rightarrow \overline{GF}= \overline{GC} \\ \Rightarrow 2\overline{EF}=a+2 \Rightarrow 2\sqrt{a^2+{1\over 4}} =a+2 \Rightarrow 4a^2+1= a^2+4a+4 \Rightarrow 3a^2-4a-3=0  \\ \Rightarrow a={2+\sqrt{13} \over 3} \Rightarrow \tan \angle AFE = {1/2 \over a} ={3 \over 4+2\sqrt{13}} ={3(2\sqrt{13}-4) \over (2\sqrt{13}-4)(2\sqrt{13}+4)} ={\sqrt{13}-2 \over 6}\\，故選 \bbox[red, 2pt]{(D)}$$

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二、複選題(64%)

解：
$$(A)\bigcirc: A=\left[ \matrix{0 & -1 \\ -1 & 0} \right] \Rightarrow A^2= \left[ \matrix{1 & 0 \\ 0 & 1} \right]=I \Rightarrow \cases{A^{91} =A^{90}\cdot A = A\\ A^{71}=A^{70} \cdot A=A} \Rightarrow A^{91}=A^{71} \\ (B)\times: B= \left[ \matrix{1/2 & -\sqrt 3/2 \\ \sqrt 3/2 & 1/2} \right] = \left[ \matrix{\cos (\pi/3) & -\sin (\pi/3) \\ \sin (\pi/3) & \cos (\pi/3)} \right] \\\qquad \Rightarrow \cases{B^{91} = \left[ \matrix{\cos (91\pi/3) & -\sin (91\pi/3) \\ \sin (91\pi/3) & \cos (91\pi/3)} \right] = \left[ \matrix{\cos (\pi/3) & -\sin (\pi/3) \\ \sin (\pi/3) & \cos (\pi/3)} \right] \\B^{71} = \left[ \matrix{\cos (71\pi/3) & -\sin (71\pi/3) \\ \sin (71\pi/3) & \cos (71\pi/3)} \right] = \left[ \matrix{\cos (\pi/3) & \sin (\pi/3) \\ -\sin (\pi/3) & \cos (\pi/3)} \right]} \Rightarrow B^{91} \ne B^{71} \\(C)\bigcirc: C= \left[ \matrix{1/2 & \sqrt  3/2 \\ \sqrt 3/2 & -1/2} \right] \Rightarrow C^2=\left[ \matrix{1 & 0 \\ 0 & 1} \right] \Rightarrow C^{91}=C^{71} =C \\(D)\bigcirc: \cases{B^5= \left[ \matrix{\cos (5\pi/3) & -\sin (5\pi/3) \\ \sin (5\pi/3) & \cos (5\pi/3)} \right] =\left[ \matrix{1/2 & \sqrt 3/2 \\ -\sqrt 3/2 & 1/2} \right] \\D =\left[ \matrix{-1 & 1 \\ -1 & -1} \right] \Rightarrow D^2= \left[ \matrix{0 & -2 \\ 2 & 0} \right] \Rightarrow D^3= \left[ \matrix{2 & 2 \\ -2 & 2} \right]} \\ \qquad \Rightarrow \cases{B^5D^3 = \left[ \matrix{1/2 & \sqrt 3/2 \\ -\sqrt 3/2 & 1/2} \right] \left[ \matrix{2 & 2 \\ -2 & 2} \right] = \left[ \matrix{-\sqrt 3+1 & \sqrt 3 +1 \\ -\sqrt 3-1 & -\sqrt 3+1} \right] \\ D^3B^5= \left[ \matrix{2 & 2 \\ -2 & 2} \right] \left[ \matrix{1/2 & \sqrt 3/2 \\ -\sqrt 3/2 & 1/2} \right] = \left[ \matrix{-\sqrt 3+1 & \sqrt 3 +1 \\ -\sqrt 3-1 & -\sqrt 3+1} \right]} \Rightarrow B^5D^3=D^3B^5  \\ 故選\bbox[red, 2pt]{(ACD)}$$

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解：
$$(B)\bigcirc: |x|+|y|+|z|=18,若其中一變數為0(假設是z),|x|+|y|=18,若x,y均大於0,\\則有(1,17),(2,16),...,(17,1),共17組解; x,y可交換,且可正可負,因此有17\times 4=68組解;\\加上可能x=0,或y=0,因此共有 68\times 3=204組解;另一類是x,y,z有兩個為0,例|x|=18,x=\pm 18;\\此類有2\times 3=6組解;最後一類是x,y,z都不為0,則x+y+z=18的正整數解有H^3_{15}=C^{17}_{15}=136 \\\Rightarrow |x|+|y|+|z|=18非0的解共有136\times 2^3=1088組解;綜合上述,共有204+6+1088 =1298組解 \\ (C)\bigcirc:\begin{array}{} 取球 & 次數\\\hline 2W1R & C^6_2C^4_1=60\\ 1W2R & C^6_1C^4_2 =36 \\ 2W1B & C^6_2C^3_1 =45 \\ 1W2B & C^6_1C^3_2 =18 \\ 2R1B & C^4_2C^3_1 = 18 \\ 1R2B & C^4_1C^3_2 =12\\\hline & \sum= 189\end{array} \Rightarrow 機率為{189\over C^{13}_3} ={189\over 286}\\ 故選\bbox[red, 2pt]{(BC)}$$

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解：
$$f(x)=(1+x-x^2)^{50}= 1+ax+bx^2+\cdots + cx^{100} \\\Rightarrow f'(x)=50(1+x-x^2)^{49}(1-2x) = a+2bx+ \cdots +100cx^{99} \\ \Rightarrow f''(x)=50\cdot 49(1+x-x^2)^{48}(1-2x)^2 -100(1+x-x^2)^{49}= 2b+ \cdots +9900x^{98}\\ \Rightarrow \cases{f'(0)= 50=a\\ f''(0)=50\times 49-100=2b \\ (-1)^{50}=c} \Rightarrow \cases{a=50 \\b =1175 \\c=1} \Rightarrow a+b+c =1226\\故選\bbox[red, 2pt]{(BD)}$$

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解：
$$(A)\bigcirc: a= p({1\over 2}+{1\over 2^2}+{1\over 2^3}+{1\over 2^4}) = {15\over 16}p \\(B)\times: 0< p<1 \Rightarrow a={15\over 16}p < {15\over 16} \Rightarrow a< {15\over 16} \\(C)\times: 總所得小於1/3的情況:(0,0,0,0),(0,0,0,1), (0,0,1,0),(0,1,0,0), (0,0,1,1),(0,1,0,1)\\ \qquad,機率和為 (1-p)^4+ 3(1-p)^3p+ 2p^2(1-p)^2 =(1-p)^2(1+p) \Rightarrow b=1-(1-p)^2(1+p)\\ \qquad = p+p^2-p^3為p的3次式 \\(D)\bigcirc: \cases{p^2-p^3=p^2(1-p) >0 \Rightarrow p+p^2-p^3 > p\\p^3>0 \Rightarrow p+p^2-p^3 < p+p^2 }\Rightarrow p < b< p+p^2\\故選\bbox[red, 2pt]{(AD)}$$

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第二部分：綜合題（共 60 分）

一、填充題（每題 4 分，共 36 分）

解：

$$\cases{a=3^x-9\\ b=3-9^x } \Rightarrow \cases{(3^x-9)^3-(3-9^x)^3 = a^3-b^3\\ (3^x+9^x-12)^3 = (a-b)^3} \Rightarrow a^3-b^3 =(a-b)^3 \\ \Rightarrow (a-b)(a^2+ab+b^2) =(a-b)^3 \Rightarrow a^2+ab+b^2=(a-b)^2=a^2-2ab +b^2\\ \Rightarrow 4ab=0 \Rightarrow \cases{a=0 \\ b=0 \\a=b} \Rightarrow \cases{3^x-9=0 \\ 3-9^x=0 \\ 3^x-9=3-9^x}\Rightarrow \cases{3^x=3^2 \\ 3=3^{2x} \\ (3^x-3)(3^x+4)=0} \\\Rightarrow \bbox[red, 2pt]{\cases{x=2 \\ x=1/2 \\ x=1}}$$

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解：

$$過A作水平線與過B點作垂直線，兩線交於F點，即ACBF為一矩形，見上圖；\\延長\overline{CE}交\overline{BF}於H點，延長\overline{CD}交\overline{AF}於G點，則\overline{BH} =\overline{HF}且\overline{AG} =\overline{GF}，說明如下:\\ \cases{{\overline{EP} \over \overline{AC}} ={1\over 3} \\ {\overline{CP} \over \overline{CB}}= {2\over 3} ={\overline{EP} \over \overline{HB}}} \Rightarrow \overline{EP} ={1\over 3}\overline{AC} ={2\over 3} \overline{HB} \Rightarrow \overline{HB}= {1\over 2}\overline{AC} ={1\over 2}\overline{FB} ;\\ 同理可求得 \overline{AG}={1\over 2}\overline{AF};$$

$$令\cases{\overline{BH} =\overline{HF}=a \\\overline{AG} =\overline{GF}=b}，則\cases{\sin \alpha = b/\overline{CG} = {b\over \sqrt{4a^2+b^2}} \\ \sin \gamma =a/\overline{CH} ={a\over \sqrt{a^2+4b^2}}} \\ 在\triangle CGH \Rightarrow \cos \beta={\overline{CG}^2+\overline{CH}^2 - \overline{GH}^2 \over 2\times \overline{CG}\times \overline{CH}} ={(4a^2+b^2)+(a^2+4b^2)-(a^2+b^2) \over 2\sqrt{(4a^2+b^2)(a^2+4b^2)}} ={2(a^2+b^2) \over \sqrt{(4a^2+b^2)(a^2+4b^2)}} \\ \Rightarrow 1-\cos^2 \beta=1-{4a^4+8a^2b^2+4b^4 \over 4a^4+17a^2b^2+4b^4} ={9a^2b^2 \over 4a^4+17a^2b^2+4b^4} \Rightarrow \sin \beta = {3ab \over \sqrt{(4a^2+b^2)(a^2+4b^2)}} \\ \Rightarrow {\sin \alpha \cdot \sin \gamma \over \sin \beta} ={ab \over 3ab} =\bbox[red, 2pt]{1\over 3}$$

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解：

$$此題相當於求兩圖形\cases{y=||x|-4| \\y=mx+3} 相交兩點的條件\\ y=||x|-4|圖形可區分成4條直線L_1,L_2,L_3,L_4，其斜率\cases{m_{L_1} = m_{L_3} =-1 \\ m_{L_2} = m_{L_4} =1};\\ 又直線y=mx+3必過P(0,3)，其斜率必須滿足\cases{ m_{\overline{PC}} > m> m_{L_3} \\ m_{\overline{PB}}< m < m_{L_2}} \Rightarrow \bbox[red, 2pt]{\cases{ -1< m < -3/4\\ 3/4 < m < 1}}$$

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解：

$$令\cases{B(0,0) \\ A(0,\sqrt 5) \\ C(2\sqrt 5,0)\\ D(a,b)}，由\cases{\overline{AD}=\sqrt {21} \\ \overline{CD}=6} \Rightarrow \cases{a^2+(b-\sqrt 5)^2 =21 \\  (a-2\sqrt 5)^2+b^2=36} \Rightarrow b=2a \\ \Rightarrow a={10+2\sqrt 5\over 5} \Rightarrow a^2= {120+40\sqrt 5 \over 25} \Rightarrow \overline{BD}^2 = a^2+b^2 =5a^2 = 24+8\sqrt 5 \\ \Rightarrow \overline{BD} =\sqrt{24+8\sqrt 5} = \bbox[red, 2pt]{2+2\sqrt 5}$$

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解：

$$令z ={1\over 2}(\cos{\pi \over 3}+i\sin{\pi \over 3}) = {1\over 2}e^{\pi i\over 3} \Rightarrow a+bi=\sum_{n=1}^\infty z^n = {z\over 1-z}= {{1\over 2}e^{\pi i\over 3} \over 1-{1\over 2}e^{\pi i\over 3}} = {{1\over 4}+{\sqrt 3\over 4}i \over {3\over 4}-{\sqrt 3\over 4}i} \\= {{4\over 16}\sqrt 3 i \over {12\over 16}} ={\sqrt 3\over 3}i  \Rightarrow (a,b) = \bbox[red, 2pt]{(0,{\sqrt 3\over 3})}$$

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解：

$$此題相當於求兩圖形\cases{y=f(x)=\sin x-3\cos x\\ y=k}，在0\le x \le \pi 限制下，有兩個交點的條件；\\ f(x)=\sin x-3\cos x = \sqrt{10}\sin (x-\theta) \Rightarrow f(x)的最大值為\sqrt{10}，最小值為-\sqrt{10};\\ 又f'(x)=\cos x+3\sin x \Rightarrow f'(x)=0 \Rightarrow \tan x = -{1\over 3} \Rightarrow \tan {5\over 6}\pi <\tan x < \tan \pi \\ 因此若要兩個交點，則 f(\pi) \le k< f的最大值 \Rightarrow \bbox[red, 2pt]{3 \le k < \sqrt{10}}$$

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7.袋中有12個白球、 8 個紅球，每次隨機抽取1球，取後不放回，直到所有球取完為止。求取球過程中至少有一次遇到取得之白球、紅球個數相等之機率為_______？

解：

在格子坐標上,拿一個白球代表向右走一步,拿一個紅球代表向上走一步;
12個白球與8個紅球要取完,相當於從(0,0)走到(12,8),共有$C^{20}_{12}$種走法;
取球過程中,紅白兩球球數曾經相等,代表從(0,0)走到(12,8),路線經過直線$x=y$;完全不經過$x=y$的路線共有$C^{20-1}_{12-1}-C^{20-1}_{12}=C^{19}_{11}-C^{19}_{12}$條;因此紅白兩球球數曾經相等的機率為
$$1-{C^{19}_{11}-C^{19}_{12} \over C^{20}_{12}}=1-{1\over 5}=\bbox[red, 2pt]{4\over 5}$$

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解：

$$假設\cases{B(0,0)\\ A(0,6)\\ C(6,0)} \Rightarrow \cases{D(3,3) \\ P(2,0) \\ Q(4,0)} \Rightarrow \cases{\overline{BD}:y=x\\ \overline{AP}:y=-3x+6 \\ \overline{AQ}: -{3\over 2}x+6} \Rightarrow \cases{E({3\over 2},{3\over 2}) \\ F({12\over 5},{12\over 5})} \\ \Rightarrow \cases{\overline{BE}={3\over 2}\sqrt 2 \\ \overline{EF}={9\over 10}\sqrt 2 \\ \overline{FD}={3\over 5}\sqrt 2} \Rightarrow \overline{BE}: \overline{EF}:\overline{FD} = {3\over 2}:{9\over 10}: {3\over 5} = 15:9:6= \bbox[red, 2pt]{5:3:2}$$

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9.設甲袋中有2白球，乙袋中有3紅球，今每次自各袋中隨機取一球作交換，趨於穩定時，甲袋中有1白球1紅球之機率為 ______？

解：

$$甲袋有3種狀態:\cases{S_1=2白\\ S_2=1白1紅\\ S_3=2紅} \Rightarrow \begin{array}{c|ccc}& &新狀態\\ & S_1 & S_2 & S_3 \\\hline 原\;S_1 & 0 & 1 & 0 \\ 狀\;S_2 & 1/6 & 1/2 & 1/3 \\ 態\;S_3 & 0 & 2/3 & 1/3\end{array} =P^T \\\Rightarrow 轉移矩陣P=\left[ \matrix{0 & 1/6 &0 \\ 1 & 1/2 & 2/3 \\ 0 & 1/3& 1/3} \right]\\ \Rightarrow 穩定後\left[\matrix{a \\ b \\ 1-a-b} \right] =\left[ \matrix{0 & 1/6 &0 \\ 1 & 1/2 & 2/3 \\ 0 & 1/3& 1/3} \right]\left[\matrix{a \\b\\ 1-a-b} \right] \Rightarrow \cases{a=0.1(S_1)\\ b=0.6 (S_2)\\ 1-a-b=0.3(S_3)} \\ \Rightarrow 穩定後甲袋有1白1紅的機率為\bbox[red, 2pt]{0.6}$$

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二、計算題（每題 8 分，共 24 分）

解：

$$x^3-2x^2-3x+1=0之三根為\alpha,\beta,\gamma \Rightarrow \cases{\alpha + \beta +\gamma = 2 \\ \alpha \beta + \beta\gamma + \gamma\alpha=-3 \\ \alpha \beta \gamma =-1} \\ 另三次方程式三根為\cases{x_1 ={\alpha-1 \over 2\alpha+3} \\ x_2 ={\beta-1 \over 2\beta+3}\\ x_3 ={\gamma-1 \over 2\gamma+3}} \\ \Rightarrow x_1x_2x_3= {(\alpha-1)(\beta-1)(\gamma-1) \over (2\alpha+3)(2\beta+3)(2\gamma+3)} = {\alpha\beta\gamma - (\alpha\beta +\beta\gamma +\gamma\alpha)+(\alpha+\beta+\gamma)-1 \over 8\alpha\beta\gamma +12 (\alpha\beta +\beta\gamma +\gamma\alpha)+ 18(\alpha+\beta+\gamma)+27 } \\ ={-1+3+2-1 \over -8-36+36 +27} ={3 \over 19}\\ x_1+x_2+x_3 ={(\alpha-1)(2\beta+3)(2\gamma+3) + (\beta-1)(2\alpha+3)(2\gamma+3) +(\gamma-1)(2\alpha+3 )(2\beta+3) \over (2\alpha+3)(2\beta+3)(2\gamma+3)} \\={ 12\alpha\beta\gamma +8(\alpha\beta+\beta\gamma +\gamma \alpha)-3(\alpha+\beta +\gamma)-27\over 19} ={-12+24-6-27 \over 19} =-{21\over 19} \\ x_1x_2 + x_2x_3 + x_3x_1= {(\alpha-1)(\beta-1) \over (2\alpha+3)(2\beta+3)} +{(\beta-1)(\gamma-1) \over (2\beta+3)(2\gamma+3)} +{(\alpha-1)(\gamma-1) \over (2\alpha+3)(2\gamma+3)} \\={ (\alpha-1)(\beta-1)(2\gamma+3) +(\beta-1)(\gamma-1)(2\alpha+3) + (\alpha-1)(\gamma-1)(2\beta+3)\over (2\alpha+3 )(2\beta+3)(2\gamma+3)} \\= {6\alpha\beta\gamma -(\alpha\beta +\beta\gamma + \gamma\alpha)-4(\alpha +\beta+\gamma)+9 \over 19} ={-6+3-8+9 \over 19} =-{2\over 19} \\ \Rightarrow 三次方程式為:\bbox[red, 2pt]{ x^3+{21\over 19}x^2-{2\over 19}x-{3\over 19}=0}$$

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解：

$$由上圖可知: 同一斜列的元素有相同的(i+j)，且該斜列有(i+j-1)個元素；\\因此(i,j)位於第1+2+\cdots + (i+j-2)+i = {(i+j-1)(i+j-2)\over 2}+i\\ \Rightarrow a_{ij} =(i+j-1)(i+j-2)+2i =\bbox[red, 2pt]{(i+j)^2-3(i+j)+2i+2}$$

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3.如圖，有一個底半徑為 5 公分的圓柱體，被一個通過直徑 AB 且與底面夾 45 角的平面所截，試求所截出的立體體積。(如圖陰影與斜線的體積)

解：

$$利用公式{2\over 3}R^3\tan \theta ={2\over 3}\times 5^3\times 1= \bbox[red, 2pt]{{250\over 3}}\\ 公式來源及證明\to \href{https://chu246.blogspot.com/2020/12/blog-post.html}{按這裡}$$

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