h) x⋮ 8; x⋮ 10; x⋮ 12 và 450 < x < 500 i) 72 ⋮ x; 60 ⋮ x và x > 6 k) x⋮ 5 và 210 < 7x < 280 l) (x+ 5) ⋮ (x +1)

Question

h) x⋮ 8; x⋮ 10; x⋮ 12 và 450 < x < 500 i) 72 ⋮ x; 60 ⋮ x và x > 6
k) x⋮ 5 và 210 < 7x < 280 l) (x+ 5) ⋮ (x +1) m) (2x + 15) ⋮ (x + 6) n) (x+ 2) + (x + 4) + (x + 6) + ….. + (x + 2008) = 1010024

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Natalia 2 năm 2021-09-09T18:45:10+00:00 1 Answers 18 views 0

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  1. tuvi
    0
    2021-09-09T18:46:29+00:00
    Reply

    Đáp án:

    $\begin{array}{l}
    h)x \vdots 8;x \vdots 10;x \vdots 12\\
     \Rightarrow x \in BC\left( {8;10;12} \right)\\
    8 = {2^3};10 = 2.5;12 = {2^2}.3\\
     \Rightarrow BCNN\left( {8;10;12} \right) = {2^3}.3.5 = 120\\
     \Rightarrow x \in \left\{ {120;240;360;480;600;…} \right\}\\
    D0:450 < x < 500\\
     \Rightarrow x = 480\\
    i)72 \vdots x;60 \vdots x\\
     \Rightarrow x \in UC\left( {72;60} \right)\\
    Do:72 = {2^3}{.3^2};60 = {2^2}.3.5\\
     \Rightarrow UCNN\left( {72;60} \right) = {2^2}.3 = 12\\
     \Rightarrow x \in \left\{ {1;2;3;4;6;12} \right\}\\
    x > 6\\
     \Rightarrow x = 12\\
    k)x \vdots 5;210 < 7x < 280\\
     \Rightarrow 30 < x < 40\\
     \Rightarrow x = 35\\
    l)\left( {x + 5} \right) \vdots \left( {x + 1} \right)\\
    x + 5 = x + 1 + 4\\
     \Rightarrow 4 \vdots \left( {x + 1} \right)\\
     \Rightarrow \left( {x + 1} \right) \in \left\{ { – 4; – 2; – 1;1;2;4} \right\}\\
     \Rightarrow x \in \left\{ { – 5; – 3; – 1;0;1;3} \right\}\\
    m)\left( {2x + 15} \right) \vdots \left( {x + 6} \right)\\
    2x + 15 = 2x + 12 + 3 = 2\left( {x + 6} \right) + 3\\
    Do:2\left( {x + 6} \right) \vdots \left( {x + 6} \right)\\
     \Rightarrow 3 \vdots \left( {x + 6} \right)\\
     \Rightarrow \left( {x + 6} \right) \in \left\{ { – 3; – 1;1;3} \right\}\\
     \Rightarrow x \in \left\{ { – 9; – 7; – 5; – 3} \right\}\\
    n)\left( {x + 2} \right) + \left( {x + 4} \right) + \left( {x + 6} \right) + … + \left( {x + 2008} \right)\\
     = \left( {x + x + … + x} \right) + \left( {2 + 4 + 6 + .. + 2008} \right)
    \end{array}$

    Có số số hạng x là:$\dfrac{{2008 – 2}}{2} + 1 = 1004$

    $\begin{array}{l}
     \Rightarrow 1010024 = 1004x + \dfrac{{\left( {2008 + 2} \right).1004}}{2}\\
     \Rightarrow 1004x + 1009020 = 1010024\\
     \Rightarrow 1004x = 1004\\
     \Rightarrow x = 1
    \end{array}$

    Vậy x=1

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