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

Toán Lớp 9: tìm `x` `x^2+x=600` `x^2-x=1001`

Toán Lớp 9: tìm x
x^2+x=600
x^2-x=1001

Comments ( 2 )

  1. Đáp án :
    x \in { -25 ; 24} 
    x \in {\frac{ \sqrt{4005} +1}2 ;{1- \sqrt{4005}}/2} 
    Giải thích :
    x^2 +x =600
    => x^2 + x – 600 =0
    => x^2 +25x -24x -600 =0 
    => x(x-24)+25(x-24)=0
    => (x-24)(x+25) =0   
    Khi đó : x -24 =0 hoặc x+25 =0
    => x =24 hoặc x= -25 
    Vậy x \in { -25 ; 24} 
    $\\$
    x^2 -x =1001
    => x^2 – 2 . x . 1/2 + 1/4  – 4005/4 = 0
    => (x-1/2)^2 =4005/4
    => x -1/2 = \sqrt{4005}/2 hoặc x -1/2 =- \sqrt{4005}/2 
    Xét x -1/2 = \sqrt{4005}/2
    => x = {\sqrt{4005} + 1}/2 
    Xét x -1/2 = -\sqrt{4005}/2 
    => x = {1- \sqrt{4005}/2
    Vậy x \in {\frac{ \sqrt{4005} +1 }2 ;{1- \sqrt{4005}}/2} 

  2. Giải đáp+Lời giải và giải thích chi tiết:
    $x^2+x=600$
    $⇔x^2+x-600=0$
    $⇔x^2-24x+25x-600=0$
    $⇔x(x-24)+25(x-24)=0$
    $⇔(x+25)(x-24)=0$
    $⇔\left[\begin{matrix}x+25=0\\x-24=0\end{matrix}\right.$
    $⇔\left[\begin{matrix}x=-25\\x=24\end{matrix}\right.$
    Vậy S={-25;24}
    ——————————
    $x^2-x=1001$
    $⇔x^2-x-1001=0$
    $⇔x^2-x+\dfrac{1}{4}-\dfrac{4005}{4}=0$
    $⇒(x-\dfrac{1}{2})^2-\dfrac{4005}{4}=0$
    $⇔(x-\dfrac{1}{2}-\dfrac{3\sqrt{445}}{2})(x-\dfrac{1}{2}+\dfrac{3\sqrt{445}}{2})=0$
    $⇔(x-\dfrac{1+3\sqrt{445}}{2})(x+\dfrac{-1+3\sqrt{445}}{2})=0$
    $⇔\left[\begin{matrix}x-\dfrac{1+3\sqrt{445}}{2}=0\\x+\dfrac{-1+3\sqrt{445}}{2}=0\end{matrix}\right.$
    $⇔\left[\begin{matrix}x=\dfrac{1+3\sqrt{445}}{2}\\x=\dfrac{1-3\sqrt{445}}{2}\end{matrix}\right.$
    Vậy S={\frac{1+3\sqrt{445}}{2};\frac{1-3\sqrt{445}}{2}}

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