Pembahasan Latihan Soal Olimpiade Matematika (6)


  1. Hitung nilai \frac{7^{502}-7^{500}-144}{7^{500}-3} = …

    PEMBAHASAN :

    \frac{7^{502}-7^{500}-144}{7^{500}-3} = \frac{7^2.7^{500}-7^{500}-144}{7^{500}-3}

    = \frac{49.7^{500}-7^{500}-49.3-3}{7^{500}-3}

    = \frac{49(7^{500}-3)-(7^{500}-3)}{7^{500}-3}

    = \frac{49(7^{500}-3)}{7^{500}-3} \frac{(7^{500}-3)}{7^{500}-3}

    = 49 – 1

    = 48

  2. Hitunglah \frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{4}} + … + \frac{1}{\sqrt{9800}+\sqrt{9801}} = …

    PEMBAHASAN :

    Kita harus merasionalan setiap pecahan dengan cara mengalikan dengan sekawannya.

    \frac{1}{1+\sqrt{2}} x \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{1-\sqrt{2}}{1-2}

    = \sqrt{2} – 1

    \frac{1}{\sqrt{2}+\sqrt{3}} x \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} = \frac{\sqrt{2}-\sqrt{3}}{2-3}

    = \sqrt{3}\sqrt{2}

    .

    .

    .

    \frac{1}{\sqrt{9800}+\sqrt{9801}} x \frac{\sqrt{9800}-\sqrt{9801}}{\sqrt{9800}-\sqrt{9801}} = \frac{\sqrt{9800}-\sqrt{9801}}{9800-9801}

    = \sqrt{9801}\sqrt{9800}

    Setelah dirasionalkan, kemudian dijumlahkan semua bilangannya

    (\sqrt{2} – 1) + (\sqrt{3}\sqrt{2}) + (\sqrt{4}\sqrt{3}) + … + (\sqrt{9801}\sqrt{9800})

    = -1 + \sqrt{9801}

    = -1 + 99

    = 98

  3. Jika diketahui a + b = 1 dan a2 + b2 = 2, berapakah a4 + b4 ?

    PEMBAHASAN :

    (a + b)2 = a2 + 2ab + b2

    2ab = (a + b)2 – (a2 + b2)

    = (1)2 – 2

    = -1

    ab = -1/2

    (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

    = a4 + 4ab(a2 + b2) + 6(ab)2 + b4

    a4 + b4 = (a + b)4 – 4ab(a2 + b2) – 6(ab)2

    = (1)4 – 4(-1/2)(2) – 6(-1/2)2

    = 1 + 4 – 3/2

    = 3 1/2

  4. Hasil pembagian dari \frac{(2n)!}{2^n.1.3.5...(2n-3)(2n-1)} adalah …

    PEMBAHASAN :

    \frac{(2n)!}{2^n.1.3.5...(2n-3)(2n-1)} = \frac{1!.2.3...(2n-5)(2n-4)(2n-3)(2n-2)(2n-1)(2n)}{2^n.1.3.5...(2n-3)(2n-1)}

    = \frac{2.4.6...(2n-4)(2n-2)(2n)}{2^n}

    = \frac{2(1).2(2).2(3)...2(n-2)2(n-1)2(n)}{2^n}

    = \frac{(2.2...2)(1.2...(n-2)(n-1)(n))}{2^n}

    = \frac{2^n.n!}{2^n}

    = n!

  5. Jika C3n = 2n, maka C72n = …

    PEMBAHASAN :

    \frac{n!}{(n-3)!.3!} = 2n

    \frac{(n-3)!(n-2)(n-1)n}{(n-3)!.3!} = 2n

    \frac{(n-2)(n-1)n}{3!} = 2n

    (n – 2)(n – 1)n = 2n.3!

    (n – 2)(n – 1) = 2.3!=

    n2 – 3n + 2 = 12

    n2 – 3n – 10 = 0

    (n – 5)(n + 2) = 0

    n = 5 atau n = -2

    untuk n = 5

    C710 = \frac{10!}{(10-7)!.7!}

    = \frac{7!.8.9.10}{3!.7!}

    = \frac{8.9.10}{3!}

    = 4.3.10

    = 120

  6. Koefesien x7 dalam (1 + x)11 adalah …

    PEMBAHASAN :

    Perhatikan segitiga pascal:

    1

    1 1

    1 2 1

    1 3 3 1

    1 4 6 4 1

    1 5 10 10 5 1

    1 6 15 20 15 6 1

    1 7 21 35 35 21 7 1

    1 8 28 56 70 56 28 8 1

    1 9 36 84 126 126 84 36 9 1

    1 10 45 120 210 252 210 120 45 10 1

    1 11 55 165 330 462 462 330 165 55 11 1 Perhatikan penjabaran : (1 + x)11 = 111 + (11)110x1 + (55)19x2 + (165)18x3 + (330)17x4 + (462)16x5 + (462)15x6 + (330)14x7 + (165)13x8 + (55)12x9 + (11)11x10 + x11 Jadi koefesien dari x7 adalah 330

  7. Jika A = \left( \begin{array}{rr} 0 & a\\ 1 & 0\end{array} \right) maka A2009 = …

    PEMBAHASAN :

    A2 = A.A = \left( \begin{array}{rr} 0 & a\\ 1 & 0\end{array} \right) \left( \begin{array}{rr} 0 & a\\ 1 & 0\end{array} \right) = \left( \begin{array}{rr} a & 0\\ 0 & a\end{array} \right)

    A3 = A2.A = \left( \begin{array}{rr} a & 0\\ 0 & a\end{array} \right) \left( \begin{array}{rr} 0 & a\\ 1 & 0\end{array} \right) = \left( \begin{array}{rr} 0 & a^2\\ a & 0\end{array} \right)

    A4 = A3.A = \left( \begin{array}{rr} 0 & a^2\\ a & 0\end{array} \right) \left( \begin{array}{rr} 0 & a\\ 1 & 0\end{array} \right) = \left( \begin{array}{rr} a^2 & 0\\ 0 & a^2\end{array} \right)

    A5 = A4.A = \left( \begin{array}{rr} a^2 & 0\\ 0 & a^2\end{array} \right) \left( \begin{array}{rr} 0 & a\\ 1 & 0\end{array} \right) = \left( \begin{array}{rr} 0 & a^3\\ a^2 & 0\end{array} \right)

    A6 = A5.A = \left( \begin{array}{rr} 0 & a^3\\ a^2 & 0\end{array} \right) \left( \begin{array}{rr} 0 & a\\ 1 & 0\end{array} \right) = \left( \begin{array}{rr} a^3 & 0\\ 0 & a^3\end{array} \right)

    Berdasarkan pola di atas,kita dapat menyimpulkan :

    A2009 = A2008.A = \left( \begin{array}{rr} a^{1004} & 0\\ 0 & a^{1004}\end{array} \right) \left( \begin{array}{rr} 0 & a\\ 1 & 0\end{array} \right) = \left( \begin{array}{rr} 0 & a^{1005}\\ a^{1004} & 0\end{array} \right)

NOTE : silahkan dikoreksi dan berikan komentar jika ada kesalahan atau masih ada keambiguan dalam penyelesaian soal-soal ini.

One comment on “Pembahasan Latihan Soal Olimpiade Matematika (6)

  1. pembahasan No.1 langkah 2 ke Langkah 3 itu tolong dicek ulang mas yah, sepertinya ada kekeliruan dalam pendistribusiannya. terima kasih postingnya.

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