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

Toán Lớp 8: Giải phương trình (x^2-2x)^2 -3x^2 +6x -4 =0

Toán Lớp 8: Giải phương trình (x^2-2x)^2 -3x^2 +6x -4 =0

Comments ( 2 )

  1. $\\$
    (x^2 – 2x)^2 – 3x^2 +6x-4=0
    ⇔ (x^2 – 2x)^2 – (3x^2 – 6x)-4=0
    ⇔ (x^2 – 2x)^2 – 3 (x^2 – 2x)-4=0
    Đặt a=x^2 – 2x
    ⇔ a^2 – 3a -4=0
    ⇔ a^2 +a – 4a – 4=0
    ⇔ (a^2 +a)-(4a+4)=0
    ⇔ a (a+1)-4(a+1)=0
    ⇔(a+1)(a-4)=0
    ⇔ \(\left[ \begin{array}{l}a+1=0\\a-4=0\end{array} \right.\) $\\$ <=> \(\left[ \begin{array}{l}a=-1\\a=4\end{array} \right.\) $\\$ <=> \(\left[ \begin{array}{l}x^2 -2x=-1\\x^2-2x=4\end{array} \right.\) $\\$ <=> \(\left[ \begin{array}{l}x^2-2x+1=0\\x^2-2x-4=0\end{array} \right.\) $\\$ <=> \(\left[ \begin{array}{l}(x-1)^2=0\\(x-1)^2=5\end{array} \right.\) $\\$ <=> \(\left[ \begin{array}{l}x-1=0\\x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{array} \right.\) $\\$ <=> \(\left[ \begin{array}{l}x=1\\x=\sqrt{5}+1\\x=-\sqrt{5}+1\end{array} \right.\) 
    Vậy pt có tập nghiệm : S={1;\sqrt{5}+1; -\sqrt{5}+1}

  2. Giải đáp:
    S={1;±\sqrt{5}+1}
    Lời giải và giải thích chi tiết:
     (x^2-2x)^2-3x^2+6x-4=0
    <=> (x^2-2x)^2-3(x^2-2x)-4=0
    Đặt t=x^2-2x
    -> t^2-3t-4=0
    <=> (t^2+t)-(4t+4)=0
    <=> t(t+1)-4(t+1)=0
    <=> (t+1)(t-4)=0
    <=>\(\left[ \begin{array}{l}t+1=0\\t-4=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}t=-1\\t=4\end{array} \right.\) 
    *)t=-1
    -> x^2-2x=-1
    <=> x^2-2x+1=0
    <=> (x-1)^2=0
    <=> x-1=0
    <=> x=1
    *) t=4
    -> x^2-2x=4
    <=> x^2-2x+1=4+1
    <=> (x-1)^2=5
    <=> x-1=±\sqrt{5}
    <=> x=±\sqrt{5}+1
    Vậy S={1;±\sqrt{5}+1}

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

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