Tìm X :b)(x-3)^2-4=0 c)x^2-2.x=24 d)(x+4)^2- (x+1).(x-1)=16 e)3.(x-1)^2-3.x.(x-5)=1 f) (6.x-2)^2+(5.x-2)^2-4.(3.x-1).(5.x-2)=0

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1. Đáp án:

   b/ $\text{$x=5$ hoặc $x=1$}$

   c/ $\text{$x=6$ hoặc $x=-4$}$

   d/ $x=-\dfrac{1}{8}$

   e/ $x=-\dfrac{2}{9}$

   f/ $x=0$

   Giải thích các bước giải:

   b/ $(x-3)^2-4=0$

   $⇔ (x-3-2)(x-3+2)=0$

   $⇔ (x-5)(x-1)=0$

   $⇔ \left[ \begin{array}{l}x=5\\x=1\end{array} \right.$

   c/ $x^2-2x=24$

   $⇔ x^2-2x-24=0$

   $⇔ x^2-6x+4x-24=0$

   $⇔ x(x-6)+4(x-6)=0$

   $⇔ (x-6)(x+4)=0$

   $⇔ \left[ \begin{array}{l}x=6\\x=-4\end{array} \right.$

   d/ $(x+4)^2-(x+1)(x-1)=16$

   $⇔ x^2+8x+16-(x^2-1)-16=0$

   $⇔ 8x+1=0$

   $⇔ x=-\dfrac{1}{8}$

   e/ $3(x-1)^2-3x(x-5)=1$

   $⇔ 3(x^2-2x+1)-3x^2+15x-1=0$

   $⇔ 3x^2-6x+3-3x^2+15x-1=0$

   $⇔ 9x+2=0$

   $⇔ x=-\dfrac{2}{9}$

   f/ $(6x-2)^2+(5x-2)^2-4(3x-1)(5x-2)=0$

   $⇔ (6x-2)^2-2(6x-2)(5x-2)+(5x-2)^2=0$

   $⇔ (6x-2-5x+2)=0$

   $⇔ x=0$

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2. b) (x – 3)² – 4 = 0

   ⇔ (x – 3)² = 4

   ⇔ \(\left[ \begin{array}{l}x-3=2\\x-3=-2\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=5\\x=1\end{array} \right.\) 

   c)x² – 2.x = 24

   ⇔ x² – 2x – 24 = 0

   ⇔ x² + 4x – 6x – 24 = 0

   ⇔ x(x + 4) – 6(x + 4) = 0

   ⇔ (x + 4)(x – 6) = 0

   ⇔ \(\left[ \begin{array}{l}x+4=0\\x-6=0\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=-4\\x=6\end{array} \right.\) 

   d) (x + 4)² – (x + 1).(x – 1) = 16

   ⇔ x² + 8x + 16 – x² + 1 = 16

   ⇔ 8x = -1

   ⇔ x = -$\frac{1}{8}$

   e) 3.(x – 1)² – 3.x.(x – 5) = 1

   ⇔ 3x² – 6x + 3 – 3x² + 15x = 1

   ⇔ 9x = -2

   ⇔ x = -$\frac{2}{9}$

   f) (6.x – 2)² + (5.x – 2)² – 4.(3.x – 1).(5.x – 2) = 0

   ⇔ (6x – 2)² – 4(3x – 1)(5x – 2) + (5x – 2)² = 0

   ⇔ (6x – 2)² – 2(5x – 2)(5x – 2) + (5x – 2)² = 0

   ⇔ (6x – 2 – 5x + 2)² = 0

   ⇔ x² = 0

   ⇔ x = 0

    

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