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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1. 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

    

   Trả lời

2. Đáp án:

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

   Trả lời