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

Toán Lớp 8: Tìm x , biết : a. 3(x-2)+ x( 2-x)=0 b. 3x^2-8x+4=0 c. 4x^3 – 12x^2 =-9x

Toán Lớp 8: Tìm x , biết :
a. 3(x-2)+ x( 2-x)=0
b. 3x^2-8x+4=0
c. 4x^3 – 12x^2 =-9x

Comments ( 2 )

  1. a. 3(x-2) + x(2-x) = 0
    <=> 3(x-2) – x(x-2) = 0
    <=> (3-x)(x-2) = 0
    <=> \(\left[ \begin{array}{l}3-x=0\\x-2=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\) 
    Vậy x∈{3;2}
    b. 3x^2 – 8x + 4 = 0
    <=> 3x^2 – 2x – 6x + 4 =0
    <=> x(3x-2) – 2(3x-2) = 0
    <=> (x-2)(3x-2) = 0
    <=> \(\left[ \begin{array}{l}x-2=0\\3x-2=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=2\\x=\frac{2}{3}\end{array} \right.\) 
    Vậy x∈{0;2/3}
    c, 4x^3 – 12x^2 = -9x
    <=> 4x^3 – 12x^2 + 9x = 0
    <=> x(4x^2 – 12x + 9) = 0
    <=> x[(2x)^2 – 2.2x.3 + 3^2] = 0
    <=> x(2x-3)^2 = 0
    <=> \(\left[ \begin{array}{l}x=0\\2x-3=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=0\\x=1,5\end{array} \right.\) 
    Vậy x∈{0;1,5}

  2. Giải đáp+Lời giải và giải thích chi tiết:
    $a)3(x-2)+ x( 2-x)=0 \\3(x-2)-x(x-2)=0\\(x-2)(3-x)=0$
    =>x=2 \ hoặc \ x=3
    $b) 3x^2-8x+4=0 \\3x^2-2x-6x+4=0\\3x(x-2)-2(x-2)=0\\(x-2)(3x-2)=0$=>x=2\ hoặc\ x=2/3
    $c) 4x^3 – 12x^2 =-9x\\4x^3 – 12x^2+9x=0\\x(4x^2-12x+9)=0\\x(2x-3)^2=0$=>x=0\ hoặc \ x=3/2

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