câu này dễ thở hơn nè Tìm `x,y,z in [4,40]` sao cho `x + y + z = 62 ; xyz = 2880`

câu này dễ thở hơn nè
Tìm `x,y,z in [4,40]` sao cho `x + y + z = 62 ; xyz = 2880`

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  1. Phân tích $2880$ ra thừa số nguyên tố:

    $2880={{2}^{6}}{{.3}^{2}}.5$

     

    Giả sử $x<y<z\,\,\,\left( 4\le x,y,z\le 40 \right)$

    Ta có các trường hợp như sau:

     

    $TH_1:\begin{cases}x=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^2\right)\\y=18\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2.3^2\right)\\z=40\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^3.5\right)\end{cases}$

     

    $TH_2:\begin{cases}x=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^2\right)\\y=20\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^2.5\right)\\z=36\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^2.3^2\right)\end{cases}$

     

    $TH_3:\begin{cases}x=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^2\right)\\y=24\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^3.3\right)\\z=30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2.3.5\right)\end{cases}$

     

    $TH_4:\begin{cases}x=5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(5\right)\\y=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^4\right)\\z=36\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^2.3^2\right)\end{cases}$

     

    $TH_5:\begin{cases}x=5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(5\right)\\y=18\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2.3^2\right)\\z=32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^5\right)\end{cases}$

     

    $TH_6:\begin{cases}x=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2.3\right)\\y=12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^2.3\right)\\z=40\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^3.5\right)\end{cases}$

     

    $TH_7:\begin{cases}x=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2.3\right)\\y=15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(3.5\right)\\z=32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^5\right)\end{cases}$

     

    $TH_8:\begin{cases}x=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2.3\right)\\y=20\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^2.5\right)\\z=24\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^3.3\right)\end{cases}$

     

    $TH_9:\begin{cases}x=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2.5\right)\\y=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2^4\right)\\z=18\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2.3^2\right)\end{cases}$

     

    Ta chỉ nhận $TH_1$ vì $x+y+z=62$

     

    Vậy có $6$ kết quả $\begin{cases}x=4\\y=18\\z=40\end{cases}$   và các hoán vị của $x,y,z$

     

     

     

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