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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: tìm x nguyên để A đặt giá trị nguyên A = x + 2 / x^2 – x +3

Toán Lớp 8: tìm x nguyên để A đặt giá trị nguyên
A = x + 2 / x^2 – x +3

Comments ( 2 )

  1. Giải đáp: $x\in\{0,1, 3, -2\}$
    Lời giải và giải thích chi tiết:
    Để $\dfrac{x+2}{x^2-x+3}\in Z$
    $\to x+2\quad\vdots\quad x^2-x+3$
    $\to (x-3)(x+2)\quad\vdots\quad x^2-x+3$
    $\to x^2-x-6\quad\vdots\quad x^2-x+3$
    $\to x^2-x+3-9\quad\vdots\quad x^2-x+3$
    $\to 9\quad\vdots\quad x^2-x+3$
    $\to x^2-x+3\in U(9)$
    Mà $x^2-x+3=(x-\dfrac12)^2+\dfrac{11}4>\dfrac{11}{4}$
    $\to x^2-x+3\in\{3, 9\}$
    Với $x^2-x+3=3\to x^2-x=0\to x(x-1)=0\to x\in\{0,1\}$
    Với $x^2-x+3=9\to x^2-x-6=0\to (x-3)(x+2)=0$
    $\to x\in\{3, -2\}$

  2. $A=\dfrac{x+2}{x^2-x+3}$
    Để $A$ nguyên
    $\Rightarrow x+2\vdots x^2-x+3\\\Rightarrow (x-3)(x+2)\vdots x^2-x+3\\\Rightarrow x^2 +2x-3x-6\vdots x^2-x+3\\\Rightarrow x^2 – x +3-9\vdots x^2-x+3\\\Rightarrow 9\vdots x^2-x+3$
    => x^2-x+3\in Ư (9)={1;-1;3;-3;9;-9}
    $x^2-x+3=x^2-2.x.\dfrac{1}{2}+(\dfrac{1}{2})^2+\dfrac{11}{4}=(x-\dfrac{1}{2})^2+\dfrac{11}{4}\ge \dfrac{11}{4}=2,75$
    =>x^2-x+3\in {3;9}
    TH1 : $x^2-x+3=3$
    $\Rightarrow x^2 -x=0\\\Rightarrow x(x-1)=0\\\Rightarrow \left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\\\Rightarrow \left[ \begin{array}{l}x=0\\x=1\end{array} \right.$
    TH2 : $x^2-x+3=9\\\Rightarrow x^2-x-6=0\\\Rightarrow x^2-3x+2x-6=0\\\Rightarrow x(x-3)+2(x-3)=0\\\Rightarrow (x-3)(x+2)=0\\\Rightarrow \left[ \begin{array}{l}x-3=0\\x+2=0\end{array} \right.\\\Rightarrow \left[ \begin{array}{l}x=3\\x=-2\end{array} \right.$
    Vậy x\in {0;1;3;-2} để A nguyên

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222-9+11+12:2*14+14 = ? ( )