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

Toán Lớp 8: tìm x a, (9x)^2 – 25 (x-1)^2 = 0 b, x^3 + 8 – (x+2) (x-4) =0

Toán Lớp 8: tìm x
a, (9x)^2 – 25 (x-1)^2 = 0
b, x^3 + 8 – (x+2) (x-4) =0

Comments ( 2 )

  1. Giải đáp:
    $a)9x^2-25(x-1)^2=0$
    $⇔(3x)^2-[5(x-1)]^2=0$
    $⇔(3x-5x+5)(3x+5x-5)=0$
    $⇔\left[ \begin{array}{l}3x-5x+5=0\\3x+5x-5=0\end{array} \right.$
    $⇔\left[ \begin{array}{l}x=\dfrac{-5}{2}\\x=\dfrac{5}{8}\end{array} \right.$
    $b)$
    $x^3 + 8 – (x+2) (x-4) =0$
    $⇔(x+2)(x^2-2x+4)- (x+2) (x-4) =0$
    $⇔(x+2)(x^2-2x+4-x+4)=0$
    $⇔(x+2)(x^2-3x+8)=0$
    $⇒\left[ \begin{array}{l}x+2=0\\x^2-3x+8=0(loại)\end{array} \right.$
    $⇒x=-2$
     $x^2-3x+8=x^2-2.\dfrac{3}{2}x+(\dfrac{3}{2})^2-(\dfrac{3}{2})^2+8=(x-\dfrac{3}{2})^2+\dfrac{23}{4}≥\dfrac{23}{4}>0$

  2. Giải đáp + Lời giải và giải thích chi tiết:
    a)
    9x^{2}-25(x-1)^{2}=0
    <=>(3x)^{2}-[5(x-1)]^{2}=0
    <=>[3x-5(x-1)][3x+5(x-1)]=0
    <=>(3x-5x+5)(3x+5x-5)=0
    <=>(-2x+5)(8x-5)=0
    <=> \(\left[ \begin{array}{l}-2x+5=0\\8x-5=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=\dfrac{5}{8}\end{array} \right.\) 
    Vậy S={(5)/(2);(5)/(8)}
    b)
    x^{3}+8-(x+2)(x-4)=0
    <=>(x+2)(x^{2}-2x+4)-(x+2)(x-4)=0
    <=>(x+2)(x^{2}-2x+4-x+4)=0
    <=>(x+2)(x^{2}-3x+8)=0
    <=> \(\left[ \begin{array}{l}x+2=0\\x^{2}-3x+8=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=-2\\x^{2}-3x+8=0\  (1)\end{array} \right.\) 
    Giải (1)
    x^{2}-3x+8=0
    <=>x^{2}-2.x.(3)/(2)+(9)/(4)+(23)/(4)=0
    <=>(x-(3)/(2))^{2}=-(23)/(4)\    (Loại)
    Vậy S={-2}
     

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