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

Toán Lớp 8: Tìm x: a, `2x^2 – 4x = 0` b, `(3x-2)^2 – 9x^2 = 0` c, `x^2 – 3x = 0` d, `(2x+1)^2 – 4x^2 = 0`

Toán Lớp 8: Tìm x:
a, 2x^2 – 4x = 0
b, (3x-2)^2 – 9x^2 = 0
c, x^2 – 3x = 0
d, (2x+1)^2 – 4x^2 = 0

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:
    Tìm x.
    a)
    2x^2-4x=0
    <=>2x.x-2x.2=0
    <=>2x(x-2)=0
    <=>[(x=0),(x-2=0):}
    <=>[(x=0),(x=2):}
    Vậy x\in{0;2}
    b)
    (3x-2)^2-9x^2=0
    <=>(3x-2)^2-(3x)^2=0
    <=>(3x-2-3x)(3x-2+3x)=0
    <=>-2.(6x-2)=0
    <=>6x-2=0
    <=>6x=2
    <=>x=1/3
    Vậy x=1/3
    c)
    x^2-3x=0
    <=>x.(x-3)=0
    <=>[(x=0),(x-3=0):}
    <=>[(x=0),(x=3):}
    Vậy x\in{0;3}
    d)
    (2x+1)^2-4x^2=0
    <=>(2x+1)^2-(2x)^2=0
    <=>(2x+1-2x)(2x+1+2x)=0
    <=>1.(4x+1)=0
    <=>4x+1=0
    <=>4x=-1
    <=>x=-1/4
    Vậy x=-1/4

  2. Giải đáp+Lời giải và giải thích chi tiết:
      a, 2x^2 – 4x = 0
    -> 2x(x-2) = 0
    -> \(\left[ \begin{array}{l}2x=0\\x-2=0\end{array} \right.\) 
    -> \(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\) 
    b, (3x-2)^2 – 9x^2 = 0
    -> (3x-2)^2 – (3x)^2 = 0
    -> (3x-2-3x)(3x-2+3x) = 0
    -> 2.(6x – 2) = 0
    -> 6x- 2 = 0
    -> 6x = 2
    -> x = 1/3
    c, x^2 – 3x = 0
    -> x(x-3)=0
    -> \(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\) 
    -> \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\) 
    d, (2x+1)^2 – 4x^2 = 0
    -> (2x+1)^2 – (2x)^2 = 0
    -> (2x + 1 – 2x)(2x+  1+ 2x) =0
    -> 1.(4x + 1 )=  0
    -> 4x + 1 =0
    -> 4x = -1
    -> x = -1/4

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

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