Tìm x: 1) 2x ² + 3. ( x-1 ) . (x+1 ) = 5x . ( x+1 ) 2) 4. ( x+3 ) . ( 3x-2 ) -3. ( x-1 ) . ( x-1) = -27

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1. Đáp án:

    $1) x=\dfrac{-3}{5}\\
   2)
    {\left[\begin{aligned}x=0\\x=\dfrac{-34}{9}\end{aligned}\right.}\\$

   Giải thích các bước giải:

    $1) 2x^2+3(x-1)(x+1)=5x(x+1)\\
   \Leftrightarrow 2x^2+3(x^2-1)=5x^2+5x\\
   \Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\\
   \Leftrightarrow -5x=3\\
   \Leftrightarrow x=\dfrac{-3}{5}\\
   2)
   4(x+3)(3x-2)-3(x-1)(x-1)=-27\\
   \Leftrightarrow 4(3x^2-2x+9x-6)-3(x^2-2x+1)=-27\\
   \Leftrightarrow 12x^2+28x-24-3x^2+6x-3=-27\\
   \Leftrightarrow 9x^2+34x-27=-27\\
   \Leftrightarrow 9x^2+34x=-27+27=0\\
   \Leftrightarrow x(9x+34)=0\\
   \Leftrightarrow {\left[\begin{aligned}x=0\\9x+34=0\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x=0\\x=\dfrac{-34}{9}\end{aligned}\right.}\\$

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2. Đáp án:

   1) 2x ² + 3.(x-1).(x+1) = 5x . (x+1)

   ⇒ 2x²+3(x²-1)=5x²+5x

   ⇒ 2x²+3x²-3=5x²+5x

   ⇒ 5x²-3=5x²+5x

   ⇒ -3=5x

   ⇒ x=$\frac{-3}{5}$ 

   2) 4.(x+3).(3x-2)-3.(x-1).(x-1) = -27

   ⇒ (4x+12)(3x-2) – 3(x²-2x+1)=-27

   ⇒ 12x²-8x+36x-24-3x²+6x-3+27=0

   ⇒ 9x²+34x=0

   ⇒ x(9x+34)=0

   TH1: x=0

   TH2: 9x+34⇒ x=$\frac{-34}{9}$ 

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