Tìm x, biết: a)4x^2-9=x(2x-3) b)(x-1)^2+x(4-x)=0 c)x(x+2)-3x-6=0

Question

Tìm x, biết:
a)4x^2-9=x(2x-3)
b)(x-1)^2+x(4-x)=0
c)x(x+2)-3x-6=0

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Peyton 3 tháng 2021-09-02T15:04:38+00:00 2 Answers 17 views 0

Answers ( )

    0
    2021-09-02T15:05:51+00:00

    a) $4x^{2}$-9=x(2x-3)

    ⇔ $(2x)^{2}$-$3^{2}$=x(2x-3)

    ⇔(2x-3)(2x+3)=x(2x-3)

    ⇔ (2x-3)(2x+3-x)=0

    ⇔ (2x-3)(x+3)=0

    2x-3=0 hoặc x+3=0

    2x=3 hoặc x=-3

    x=$\frac{3}{2}$ hoặc x=-3

    b) $(x-1)^{2}$+x(4-x)=0

    ⇔ $x^{2}$-2x+1+4x-$x^{2}$=0

    ⇔ 2x+1=0

    ⇔x=$\frac{-1}{2}$

    c) x(x+2)-3x-6=0

    ⇔ x(x+2)+3(x+2)=0

    ⇔ (x+2)(x+3)=0

    x+2=0 hoặc x+3=0

    x=-2hoặc x=-3

     

    0
    2021-09-02T15:06:27+00:00

    Đáp án:

    $\begin{array}{l}
    a)4{x^2} – 9 = x\left( {2x – 3} \right)\\
     \Rightarrow {\left( {2x} \right)^2} – {3^2} = x\left( {2x – 3} \right)\\
     \Rightarrow \left( {2x – 3} \right)\left( {2x + 3} \right) – x\left( {2x – 3} \right) = 0\\
     \Rightarrow \left( {2x – 3} \right)\left( {2x + 3 – x} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    2x – 3 = 0\\
    x + 3 = 0
    \end{array} \right. \Rightarrow \left[ \begin{array}{l}
    x = \frac{3}{2}\\
    x =  – 3
    \end{array} \right.\\
    b){\left( {x – 1} \right)^2} + x\left( {4 – x} \right) = 0\\
     \Rightarrow {x^2} – 2x + 1 + 4x – {x^2} = 0\\
     \Rightarrow 2x + 1 = 0\\
     \Rightarrow x = \frac{{ – 1}}{2}\\
    c)x\left( {x + 2} \right) – 3x – 6 = 0\\
     \Rightarrow x\left( {x + 2} \right) – 3\left( {x + 2} \right) = 0\\
     \Rightarrow \left( {x + 2} \right)\left( {x – 3} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    x =  – 2\\
    x = 3
    \end{array} \right.
    \end{array}$

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