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

Toán Lớp 8: Tìm x, biết: (x+2)³ + (2x-3)³ = 0

Toán Lớp 8: Tìm x, biết:
(x+2)³ + (2x-3)³ = 0

Comments ( 2 )

  1. ~rai~
    \((x+2)^3+(2x-3)^3=0\\\Leftrightarrow [x^3+3x^2.2+3x.2^2+2^3]+[(2x)^3-3.(2x)^2.3+3.2x.3^2-3^3]=0\\\Leftrightarrow (x^3+6x^2+12x+8)+(8x^3-36x^2+54x-27)=0\\\Leftrightarrow x^3+6x^2+12x+8+8x^3-36x^2+54x-27=0\\\Leftrightarrow 9x^3-30x^2+66x-19=0\\\Leftrightarrow 9x^3-3x^2-27x^2+9x+57x-19=0\\\Leftrightarrow (9x^3-3x^2)-(27x^2-9x)+(57x-19)=0\\\Leftrightarrow 3x^2(3x-1)-9x(3x-1)+19(3x-1)=0\\\Leftrightarrow (3x-1)(3x^2-9x+19)=0\\\Leftrightarrow 3x-1=0\quad\text{(vì }3x^2-9x+19=3\left(x-\dfrac{3}{2}\right)^2+\dfrac{49}{4}>0\quad\forall x)\\\Leftrightarrow x=\dfrac{1}{3}.\\\text{Vậy }x=\dfrac{1}{3}.\)

  2. Giải đáp+Lời giải và giải thích chi tiết:
    $(x+2)^3+(2x-3)^3=0$
    $⇔(x+2)^3=-(2x-3)^3$
    $⇔x+2=3-2x$
    $⇔3x=1$
    $⇔x=\dfrac{1}{3}$
    Vậy $x=\dfrac{1}{3}$
     

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

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