Toán Lớp 8: 2x ² -x=3-6x x ²-1 = 2013(x-1)

Question

Toán Lớp 8: 2x ² -x=3-6x
x ²-1 = 2013(x-1), hướng dẫn giải giúp em bài này ạ, em cảm ơn thầy cô và các bạn nhiều.

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An Kim 2 tuần 2022-04-13T01:28:30+00:00 2 Answers 0 views 0

TRẢ LỜI ( 2 )

  1. tongtung
    0
    2022-04-13T01:29:37+00:00
    Reply

    2x²–x= 3–6x
    =>x(2x–1)=–3(1–2x)
    =>x(2x–1)= –3(1–2x)
    =>x(2x–1)+3(2x–1)=0
    => (2x–1).(x+3)=0
    => 2x–1= 0 hoặc x+3=0
    => 2x = 1 hoặc x = –3
    => x= 1/2 hoặc x = –3
    Vậy x= 1/2 hoặc x= –3

    b) x²–1= 2013(x–1)
    =>(x+1)(x–1)= 2013(x–1)
    => (x+1)(x–1)–2013(x–1)=0
    =>(x+1–2013).(x–1)= 0
    =>(x+1–2013)= 0 hoặc x–1= 0
    => x–2012= 0 hoặc x–1= 0
    => x = 2012 hoặc x = 1
    Vậy x= 2012 hoặc x= 1

  2. hatuanlan
    0
    2022-04-13T01:30:21+00:00
    Reply

    a) 2x² – x = 3 – 6x

    ⇔ 2x² – x + 6x – 3 = 0

    ⇔ x ( 2x – 1 ) + 3 ( 2x – 1 ) = 0

    ⇔ ( 2x – 1 ) ( x + 3 ) = 0

    ⇔ $\left[\begin{matrix} 2x – 1 = 0 \\  x + 3 = 0 \end{matrix}\right.$

    ⇔ $\left[\begin{matrix} 2x = 1 \\ x = 0 – 3 \end{matrix}\right.$

    ⇔ $\left[\begin{matrix} x = \dfrac{1}{2} \\ x = -3 \end{matrix}\right.$

     Vậy x = 1/2 hoặc x = -3

    ___________________________________________________

    b) x² – 1 = 2013 ( x – 1 )

    ⇔ x² – 1 – 2013 ( x – 1 ) = 0

    ⇔ ( x – 1 ) ( x + 1 ) – 2013 ( x – 1 ) = 0

    ⇔ ( x – 1 ) ( x + 1 – 2013 ) = 0

    ⇔ ( x – 1 ) ( x – 2012 ) = 0

    ⇔ $\left[\begin{matrix} x – 1 = 0 \\ x – 2012 = 0 \end{matrix}\right.$

    ⇔ $\left[\begin{matrix} x = 1 \\ x = 2012\end{matrix}\right.$

     Vậy x = 1 hoặc x = 2012

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