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

Toán Lớp 8: a) -1/2x^2 (2x^3-4x+3) b)3x.(x-2)+2.(x-2)=0 c)x.(2x-1)+4x=2 giúp mình với ạ:(

Toán Lớp 8: a) -1/2x^2 (2x^3-4x+3)
b)3x.(x-2)+2.(x-2)=0
c)x.(2x-1)+4x=2
giúp mình với ạ:(

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    a) -1/[2x^2] (2x^3 -4x+3)
    =-x+2/x -3/[2x^2]
    b)3x.(x-2)+2.(x-2)=0
    =>(x-2)(3x+2)=0
    =>\(\left[ \begin{array}{l}x-2=0\\3x+2=0\end{array} \right.\) $\\$=>\(\left[ \begin{array}{l}x=2\\x=-\dfrac{2}{3}\end{array} \right.\) 
    c)x.(2x-1)+4x=2
    =>x(2x-1)+4x-2=0
    =>x(2x-1)+2(2x-1)=0
    =>(2x-1)(x+2)=0
    =>\(\left[ \begin{array}{l}2x-1=0\\x+2=0\end{array} \right.\) $\\$=>\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-2\end{array} \right.\) 
    (Bài tham khảo)

  2. a) \frac{-1}{2}x^2 ( 2x^3 – 4x + 3 )
    = -x^5 + 2x^3 – \frac{3}{2}
      b) 3x ( x – 2 ) + 2 ( x – 2 ) =0
    ⇒ ( x – 2 ) ( 3x + 2 ) =0
    ⇒ \(\left[ \begin{array}{l}x-2=0\\3x+2=0\end{array} \right.\) 
    ⇒ \(\left[ \begin{array}{l}x=2\\x=-\dfrac{2}{3}\end{array} \right.\)
    Vậy x ∈ { 2 ; \frac{-2}{3} }
    c) x ( 2x – 1 ) + 4x = 2
    ⇒ x ( 2x – 1 ) + 4x – 2 =0
    ⇒ x ( 2x – 1 ) + 2 ( 2x – 1 )=0
    ⇒ ( x + 2 ) ( 2x – 1 )=0
    ⇒ \(\left[ \begin{array}{l}x+2=0\\2x-1=0\end{array} \right.\) 
    ⇒ \(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{2}\end{array} \right.\) 
    Vậy x ∈ { -2 ; \frac{1}{2} }

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

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