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

Toán Lớp 8: Tìm x biết: $3x^4 – 2x^3 – 53x^2 – 4x + 12=0$

Tháng Mười Hai 30, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Tìm x biết: $3x^4 – 2x^3 – 53x^2 – 4x + 12=0$

Comments ( 1 )

  1. diemmyhoa
    diemmyhoa
    30 Tháng Mười Hai, 2022 at 3:22 sáng
    Reply
    Giải đáp:
    x\in {{-\sqrt{97}-13}/6;{\sqrt{97}-13}/6;
    {\sqrt{17}+5}/2;{-\sqrt{17}+5}/2}
    Lời giải và giải thích chi tiết:
    Ta có:
    \qquad 3x^4-2x^3-53x^2-4x+12=0
    <=>3x^4-15x^3+13x^3+6x^2+6x^2-65x^2+26x-30x+12=0
    <=>(3x^4+13x^3+6x^2)-(15x^3+65x^2+30x)+(6x^2+26x+12)=0
    <=>x^2(3x^2+13x+6)-5x(3x^2+13x+6)+2(3x^2+13x+6)=0
    <=>(3x^2+13x+6)(x^2-5x+2)=0 
    <=>[3(x^2+2x. {13}/6+({13}/6)^2)-{97}/{12}].\ [ (x^2-2x. 5/2+(5/2)^2)-{17}/4]=0
    <=>[3(x+{13}/6)^2-{97}/{12}].\ [(x-5/2)^2-{17}/4]=0
    <=>$\left[\begin{array}{l}3(x+\dfrac{13}{6})^2-\dfrac{97}{12}=0\\(x-\dfrac{5}{2})^2-\dfrac{17}{4}=0\end{array}\right.$
    <=>$\left[\begin{array}{l}(x+\dfrac{13}{6})^2=\dfrac{97}{36}\\(x-\dfrac{5}{2})^2=\dfrac{17}{4}\end{array}\right.$
    <=>$\left[\begin{array}{l}x+\dfrac{13}{6}=\sqrt{\dfrac{97}{36}}=\dfrac{\sqrt{97}}{6}\\x+\dfrac{13}{6}=-\sqrt{\dfrac{97}{36}}=\dfrac{-\sqrt{97}}{6}\\x-\dfrac{5}{2}=\sqrt{\dfrac{17}{4}}=\dfrac{\sqrt{17}}{2}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{17}{4}}=\dfrac{-\sqrt{17}}{2}\end{array}\right.$
    <=>$\left[\begin{array}{l}x=\dfrac{\sqrt{97}-13}{6}\\x=\dfrac{-\sqrt{97}-13}{6}\\x=\dfrac{\sqrt{17}+5}{2}\\x=\dfrac{-\sqrt{17}+5}{2}\end{array}\right.$
    Vậy các giá trị của x thỏa mãn đề bài là:
    x\in {{-\sqrt{97}-13}/6;{\sqrt{97}-13}/6;
    \qquad {\sqrt{17}+5}/2;{-\sqrt{17}+5}/2}

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

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