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

Toán Lớp 8: Tìm x. a/ ( 4x – 3 )^2 – 3x( 3 – 4x) = 0 b/ ( 2x + 1 ) ( 4x^2 – 2x + 1 ) – 8x ( x^2 + 2 ) = 17

Toán Lớp 8: Tìm x.
a/ ( 4x – 3 )^2 – 3x( 3 – 4x) = 0
b/ ( 2x + 1 ) ( 4x^2 – 2x + 1 ) – 8x ( x^2 + 2 ) = 17

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    a)(4x-3)^2-3x(3-4x)=0
    <=>16x^2−24x+9−9x+12x^2=0
    <=>28x^2−33x+9=0
    <=>28x^2−21x-12x+9=0
    <=>(28x^2−21x)+(-12x+9)=0
    <=>7x(4x−3)−3(4x−3)=0
    <=>(4x−3)(7x−3)=0
    <=>\(\left[ \begin{array}{l}4x−3=0⇔x=\frac{3}{4}\\7x−3=0⇔x=\frac{3}{7}\end{array} \right.\)
    Vậy x∈{3/4;3/7}
     
    b)(2x+1)(4x^2-2x+1)-8x(x^2+2)=17
    <=>8x^3+4x^2−4x^2−2x+2x+1-8x^3−16x=17
    <=>1−16x=17
    <=>-16x=16
    <=>x=-1
    Vậy x=-1

  2. Giải đáp:
    ↓↓
    Lời giải và giải thích chi tiết:
    $a)$
    $(4x-3)^2-3x(3-4x)=0$
    $⇔16x^2−24x+9−9x+12x^2=0$
    $⇔28x^2−33x+9=0$
    $⇔28x^2−21x-12x+9=0$
    $⇔(28x^2−21x)+(-12x+9)=0$
    $⇔7x(4x−3)−3(4x−3)=0$
    $⇔(4x−3)(7x−3)=0$
    $⇔$\(\left[ \begin{array}{l}4x−3=0\\7x−3=0\end{array} \right.\)$⇔$\(\left[ \begin{array}{l}x=\frac{3}{4}\\x=\frac{3}{7}\end{array} \right.\)
    Vậy $x∈{3/4;3/7}$
    $b)$
    $(2x+1)(4x^2-2x+1)-8x(x^2+2)=17$
    $⇔8x^3+4x^2−4x^2−2x+2x+1-8x^3−16x=17$
    $⇔1−16x=17$
    $⇔-16x=16$
    $⇔x=-1$
    Vậy $x=-1.$

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

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