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Q. Terlampir![tex]\bold{\underline{Rules \: :}}[/tex][tex]✎ \: No \: Calcu \: ☑[/tex][tex]✎\:

Berikut ini adalah pertanyaan dari ArtX1 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Q. Terlampir!\bold{\underline{Rules \: :}}
✎ \: No \: Calcu \: ☑
✎\: No \: bahasa \: alien \: ☑︎
✎ \: No \: Jawab \: Dikomen \: ☑︎
✎ \: Memakai \: Cara \: ☑︎​
Q. Terlampir![tex]\bold{\underline{Rules \: :}}[/tex][tex]✎ \: No \: Calcu \: ☑[/tex][tex]✎\: No \: bahasa \: alien \: ☑︎ [/tex][tex]✎ \: No \: Jawab \: Dikomen \: ☑︎ [/tex][tex]✎ \: Memakai \: Cara \: ☑︎[/tex]​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawab:

99/8

Penjelasan :

Catatan : liat jawaban di web atau liat di gambar

x+y = 1, x^2+y^2 = 2\to x = \sqrt{2}\cos(t), y = \sqrt{2}\sin(t)\\\sqrt{2} (\cos(t)+\sin(t)) = 1\\ \cos(t-\tfrac{\pi}{4}) = \dfrac{1}{2}\to t = \dfrac{7\pi}{12}\cup -\dfrac{\pi}{12}\\x_1 = \sqrt{2} \cdot \cos(\tfrac{7\pi}{12}), y_1= \sqrt{2} \cdot \sin(\tfrac{7\pi}{12})\\x_1 = \dfrac{1-\sqrt{3} }{2}, y_1 = \dfrac{1+\sqrt{3}}{2} = -\dfrac{1}{1-\sqrt{3}} = -\dfrac{1}{2x_1}

jika x1 dan y1 diubah menjadi persamaan kuadrat :

(2t-(1+\sqrt{3}))(2t - (1-\sqrt{3})) = 4t^2 - 4t -2 = 0\to t^2 = t+\dfrac{1}{2}, \dfrac{1}{t} = 2t-2\\-\dfrac{1}{2t}= 1 - t\to \dfrac{1}{4t^2} = \dfrac{1}{2}-\dfrac{1}{2t} = \dfrac{3}{2}-t

t^2 = t+\dfrac{1}{2} = x^2, \dfrac{1}{t} = 2t-2\\-\dfrac{1}{2t}= 1 - t\to \dfrac{1}{4t^2} = \dfrac{1}{2}-\dfrac{1}{2t} = \dfrac{3}{2}-t = y^2

x^2+y^2 = t+\dfrac{1}{2}+\dfrac{3}{2}-t = 2\\

t^3 = t^2+\dfrac{t}{2} = t+\dfrac{1}{2}+\dfrac{t}{2} = \dfrac{3}{2}t +\dfrac{1}{2}\\\left(-\dfrac{1}{2t}\right)^3=\dfrac{1}{2}-\dfrac{3}{4t} = \dfrac{1}{2} - \dfrac{3}{2}(t-1) = 2 - \dfrac{3}{2}t\\\\t^4 = \dfrac{3t^2+t}{2} = 2t + \dfrac{3}{4}\\\left(-\dfrac{1}{2t}\right)^4=\dfrac{3}{4}-\dfrac{1}{t} = \dfrac{3}{4} - (2t-2) = \dfrac{11}{4}-2t\\t^5 = 2t^2+\dfrac{3}{4}t = \dfrac{11}{4}t+1\\\left(-\dfrac{1}{2t}\right)^5= \dfrac{15-11t}{4}

a_n = \{1,1, \dfrac{3}{2},2,\dfrac{11}{4}\}, b_n = \{(0,0),(\dfrac{1}{2},\dfrac{3}{2}),(\dfrac{1}{2},2),(\dfrac{3}{4},\dfrac{11}{4}), (1,\dfrac{15}{4})\}

a_n = b_{n-1}(y), n > 2

b_n(x) = \dfrac{a_{n-1}}{2}, b_n(y) = a_n + b_n(x) = \dfrac{2a_n+a_{n-1}}{2}\\b_n(x)+b_n(y) = x^n+y^n = a_n+a_{n-1}\\b_{11}(x)+b_{11}(y) = x^{11}+y^{11} = a_{11}+a_{10} = b_9(y)+b_8(y)\\x^{11}+y^{11} = \dfrac{2a_9+a_8+2a_8+a_7}{2}= \dfrac{2b_8(y)+3b_7(y)+b_6(y)}{2}\\x^{11}+y^{11} =\dfrac{2a_8+a_7+\tfrac{1}{2}(6a_7+3a_6+2a_6+a_5)}{2} \\ =\dfrac{2b_7(y)+b_6(y)}{2} + \dfrac{6b_6(y)+5b_5(y)+b_4(y)}{4} = \dfrac{4b_7(y)+8b_6(y)+\tfrac{86}{4}}{4}\\

x^{11}+y^{11} = \dfrac{2a_7+a_6}{2} + \dfrac{43}{8} = \dfrac{2b_6(y)+b_5(y)}{2}+\dfrac{43}{8} = \dfrac{2a_6+a_5+\tfrac{15}{4}}{2}+\dfrac{43}{8}\\= b_5(y) + \dfrac{69}{8} \; \to \text{(Lucky number)}

\Huge{\boxed{\boxed{\boldsymbol{x^{11}+y^{11} = \dfrac{15}{4}+\dfrac{69}{8} = \dfrac{99}{8}} }}}

Jawab:99/8Penjelasan :Catatan : liat jawaban di web atau liat di gambar [tex]x+y = 1, x^2+y^2 = 2\to x = \sqrt{2}\cos(t), y = \sqrt{2}\sin(t)\\\sqrt{2} (\cos(t)+\sin(t)) = 1\\ \cos(t-\tfrac{\pi}{4}) = \dfrac{1}{2}\to t = \dfrac{7\pi}{12}\cup -\dfrac{\pi}{12}\\x_1 = \sqrt{2} \cdot \cos(\tfrac{7\pi}{12}), y_1= \sqrt{2} \cdot \sin(\tfrac{7\pi}{12})\\x_1 = \dfrac{1-\sqrt{3} }{2}, y_1 = \dfrac{1+\sqrt{3}}{2} = -\dfrac{1}{1-\sqrt{3}} = -\dfrac{1}{2x_1}[/tex]jika x1 dan y1 diubah menjadi persamaan kuadrat :[tex](2t-(1+\sqrt{3}))(2t - (1-\sqrt{3})) = 4t^2 - 4t -2 = 0\to t^2 = t+\dfrac{1}{2}, \dfrac{1}{t} = 2t-2\\-\dfrac{1}{2t}= 1 - t\to \dfrac{1}{4t^2} = \dfrac{1}{2}-\dfrac{1}{2t} = \dfrac{3}{2}-t[/tex][tex]t^2 = t+\dfrac{1}{2} = x^2, \dfrac{1}{t} = 2t-2\\-\dfrac{1}{2t}= 1 - t\to \dfrac{1}{4t^2} = \dfrac{1}{2}-\dfrac{1}{2t} = \dfrac{3}{2}-t = y^2[/tex][tex]x^2+y^2 = t+\dfrac{1}{2}+\dfrac{3}{2}-t = 2\\[/tex][tex]t^3 = t^2+\dfrac{t}{2} = t+\dfrac{1}{2}+\dfrac{t}{2} = \dfrac{3}{2}t +\dfrac{1}{2}\\\left(-\dfrac{1}{2t}\right)^3=\dfrac{1}{2}-\dfrac{3}{4t} = \dfrac{1}{2} - \dfrac{3}{2}(t-1) = 2 - \dfrac{3}{2}t\\\\t^4 = \dfrac{3t^2+t}{2} = 2t + \dfrac{3}{4}\\\left(-\dfrac{1}{2t}\right)^4=\dfrac{3}{4}-\dfrac{1}{t} = \dfrac{3}{4} - (2t-2) = \dfrac{11}{4}-2t\\t^5 = 2t^2+\dfrac{3}{4}t = \dfrac{11}{4}t+1\\\left(-\dfrac{1}{2t}\right)^5= \dfrac{15-11t}{4}[/tex][tex]a_n = \{1,1, \dfrac{3}{2},2,\dfrac{11}{4}\}, b_n = \{(0,0),(\dfrac{1}{2},\dfrac{3}{2}),(\dfrac{1}{2},2),(\dfrac{3}{4},\dfrac{11}{4}), (1,\dfrac{15}{4})\}[/tex][tex]a_n = b_{n-1}(y), n > 2[/tex][tex]b_n(x) = \dfrac{a_{n-1}}{2}, b_n(y) = a_n + b_n(x) = \dfrac{2a_n+a_{n-1}}{2}\\b_n(x)+b_n(y) = x^n+y^n = a_n+a_{n-1}\\b_{11}(x)+b_{11}(y) = x^{11}+y^{11} = a_{11}+a_{10} = b_9(y)+b_8(y)\\x^{11}+y^{11} = \dfrac{2a_9+a_8+2a_8+a_7}{2}= \dfrac{2b_8(y)+3b_7(y)+b_6(y)}{2}\\x^{11}+y^{11} =\dfrac{2a_8+a_7+\tfrac{1}{2}(6a_7+3a_6+2a_6+a_5)}{2} \\ =\dfrac{2b_7(y)+b_6(y)}{2} + \dfrac{6b_6(y)+5b_5(y)+b_4(y)}{4} = \dfrac{4b_7(y)+8b_6(y)+\tfrac{86}{4}}{4}\\[/tex][tex]x^{11}+y^{11} = \dfrac{2a_7+a_6}{2} + \dfrac{43}{8} = \dfrac{2b_6(y)+b_5(y)}{2}+\dfrac{43}{8} = \dfrac{2a_6+a_5+\tfrac{15}{4}}{2}+\dfrac{43}{8}\\= b_5(y) + \dfrac{69}{8} \; \to \text{(Lucky number)}[/tex][tex]\Huge{\boxed{\boxed{\boldsymbol{x^{11}+y^{11} = \dfrac{15}{4}+\dfrac{69}{8} = \dfrac{99}{8}} }}}[/tex]

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Last Update: Thu, 22 Sep 22
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