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

Toán Lớp 8: Tìm x, biết: a) 4x(x + 1) + (3 – 2x)(3 + 2x) = 15 b) 3x(x – 20012) – x + 20012 = 0

Toán Lớp 8: Tìm x, biết:
a) 4x(x + 1) + (3 – 2x)(3 + 2x) = 15
b) 3x(x – 20012) – x + 20012 = 0

Comments ( 2 )

  1. Giải đáp:
     a, 4x(x + 1) + (3 – 2x)(3 + 2x) = 15
    ⇔ 4x^2 + 4x + 3^2 – (2x)^2 = 15
    ⇔ 4x^2 + 4x + 9 – 4x^2 = 15
    ⇔ (4x^2 – 4x^2) + 4x + 9 = 15
    ⇔ 4x + 9 = 15
    ⇔ 4x = 15 – 9 = 6
    ⇒ x = 6 : 4
    ⇒ x = 3/2
    Vậy x = 3/2
    b, 3x(x – 20012) – x + 20012 = 0
    ⇔ 3x(x – 20012) – (x – 20012) = 0
    ⇔ (x – 20012)(3x – 1) = 0
    ⇒ $\left[\begin{matrix} x – 20012 = 0\\ 3x – 1 = 0\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x = 20012\\ 3x = 1\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x = 20012\\ x = \dfrac{1}{3}\end{matrix}\right.$
    Vậy x = 20012 hoặc x = 1/3

  2. Giải đáp:
    a. 4x(x+1)+(3-2x)(3+2x)=15
    <=> 4x^2+4x+(9-4x^2)=15
    <=> 4x^2+4x+9-4x^2=15
    <=> 4x+9=15
    <=> 4x=6
    <=> x=3/2
    Vậy S={3/2}
    b. 3x(x-20012)-x+20012=0
    <=> 3x(x-20012)-(x-20012)=0
    <=> (x-20012)(3x-1)=0
    <=>\(\left[ \begin{array}{l}x-20012=0\\3x-1=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=20012\\x=\dfrac{1}{3}\end{array} \right.\) 
    Vậy S={20012;1/3}

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

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