a)5.(2x-1)=8x+1 b) I8x-3I-4x=5 – c) $\frac{x+2}{x-2}$ – $\frac{1}{x}$ = $\frac{2}{x^2-2x}$

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1. a) 5(2x – 1) = 8x + 1

   ⇔ 10x – 5 = 8x + 1

   ⇔ 10x – 8x = 1 + 5

   ⇔ 2x = 6

   ⇔ x = 3

   b)  |8x – 3| – 4x = 5

   ⇔ |8x – 3| = 5 + 4x 

   Nếu 8x – 3 ≥ 0 ⇔ 8x = 3 ⇔ x = $\frac{3}{8}$

   ⇔ 8x – 3 = 5 + 4x

   ⇔ 8x – 4x = 5 + 3

   ⇔ 4x = 8

   ⇔ x = 2

   Nếu 8x – 3 < 0 ⇔ 8x = 3 ⇔ x = $\frac{3}{8}$

   ⇔ 8x – 3 = -5 – 4x

   ⇔ 8x + 4x = -5 + 3

   ⇔ 12x = -2

   ⇔ x = $\frac{-1}{6}$

   c) $\frac{x + 2}{x – 2}$ – $\frac{1}{x}$ = $\frac{2}{x² – 2x}$

   ⇔ $\frac{x(x + 2)}{x(x – 2)}$ – $\frac{x + 2}{x(x – 2)}$ = $\frac{2}{x(x – 2)}$

   ⇔ $x^{2}$ + 2x – x – 2 = 2

   ⇔ x² + x = 0

   ⇔ x(x+1) = 0

   ⇔ x = 0 hay x + 1= 0

   ⇔ x = 0 hay x = -1

    

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2. Đáp án:

    ở dưới

   Giải thích các bước giải:

   a)5.(2x-1)=8x+1

   ⇔10x – 5 = 8x + 1

   ⇔2x = 6

   ⇔x = 1/3

   b) |8x-3|-4x=5 

   ⇔|8x-3| = 5 + 4x

   Nếu 8x-3≥0 ⇔ x ≥3/8

   ⇒8x-3=5+4x

   ⇔4x=8

   ⇔x=2(n)

   Nếu 8x-3<0⇔x<3/8

   ⇒8x-3=-5-4x

   ⇔12x=-2

   ⇔x=-1/6(n)

   c)⇔$\frac{x(x+2)}{x(x-2)}-$ $\frac{x+2}{x(x-2)}=$ $\frac{2}{x(x-2)}$ 

   ⇔x²+2x – x – 2 = 2

   ⇔x² + x = 0

   ⇔x (x+1) = 0

   ⇔x = 0 hay x = -1

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