Toán Lớp 8: Tìm x: a) -3x^2+5x+2=0 b) 4x^2–1-x(2x+1)=0

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Toán Lớp 8: Tìm x: a) -3x^2+5x+2=0
b) 4x^2–1-x(2x+1)=0, hướng dẫn giải giúp em bài này ạ, em cảm ơn thầy cô và các bạn nhiều.

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Nguyệt 1 tháng 2022-12-22T05:07:48+00:00 2 Answers 0 views 0

TRẢ LỜI ( 2 )

  1. huongtructram
    0
    2022-12-22T05:09:01+00:00
    Reply
    a)-3x²+6x-1x+2=0
    ⇔-3x(x-2)-(x-2)=0
    ⇔(x-3)(-3x-1)=0
    ⇔\(\left[ \begin{array}{l}x=3\\x=-1/3\end{array} \right.\)
    b)4x²-1-x(2x+1)=0
    ⇔4x²-1-2x²-x=0
    ⇔2x²-x-1=0
    ⇔2x²+x-2x-1=0
    ⇔x(2x+1)-(2x+1)=0
    ⇔(2x+1)(x-1)=0
    ⇔\(\left[ \begin{array}{l}x=-1/2\\x=1\end{array} \right.\)
    #NoCopy

  2. hiennhi98
    0
    2022-12-22T05:09:06+00:00
    Reply
    Hướng dẫn trả lời:
    a) -3x^2 + 5x + 2 = 0
    → -(3x^2 – 5x – 2) = 0
    → -(3x^2 – 6x + x – 2) = 0
    → -[3xcdot(x – 2) + (x – 2)] = 0
    → -(3x + 1)cdot(x – 2) = 0
    → \(\left[ \begin{array}{l}3x + 1 = 0\\x – 2 = 0\end{array} \right.\) 
    → \(\left[ \begin{array}{l}3x = – 1\\x = 2\end{array} \right.\) 
    → \(\left[ \begin{array}{l}x = \dfrac{- 1}{3}\\x = 2\end{array} \right.\) 
    Vậy x = -1/3 hoặc x = 2
    b) 4x^2 – 1 – xcdot(2x + 1) = 0
    → (2x)^2 – 1^2 – xcdot(2x + 1) = 0
    → (2x – 1)cdot(2x + 1) – xcdot(2x + 1) = 0
    → [(2x – 1) – x]cdot(2x + 1) = 0
    → (2x – 1 – x)cdot(2x + 1) = 0
    → (x – 1)cdot(2x + 1) = 0
    → \(\left[ \begin{array}{l}x – 1 = 0\\2x + 1 = 0\end{array} \right.\) 
    → \(\left[ \begin{array}{l}x = 1\\2x = -1\end{array} \right.\) 
    → \(\left[ \begin{array}{l}x = 1\\x = \dfrac{-1}{2}\end{array} \right.\) 
    Vậy x = 1 hoặc x = -1/2
    Giải đáp:
    a) x = -1/3 hoặc x = 2
    b) x = 1 hoặc x = -1/2

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