## a) 2/3 – 1/3 (x – 3/2) – 1/2 (2x + 1) = 5 b) 1/3x + 2/5 (x – 1) = 0

Question

a) 2/3 – 1/3 (x – 3/2) – 1/2 (2x + 1) = 5
b) 1/3x + 2/5 (x – 1) = 0

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1 year 2020-11-26T18:00:11+00:00 2 Answers 151 views 0

1. a) 2/3 – 1/3 ( x – 3/2 ) – 1/2 ( 2x + 1 ) = 5

=> 1/3x – 1/3 . 3/2 – 1/2 . 2x – 1/2 = 2/3 – 5 = -13/3

=> 1/3x – 1/2 – 1/2 . 2x – 1/2 = -13/3

=> 1/3x – 1/2 . 2x  = -13/3 + 1/2 + 1/2 = -10/3

=> x . ( 1/3 – 1/2 . 2 ) = -10/3

=> x . 4/3 = -10/3

=> x = -5/2

b) 1/3x + 2/5 ( x – 1 ) = 0

=> 1/3x + 2/5x – 2/5 = 0

=> x . ( 1/3 + 2/5 ) = 2/5

=> x . 11/15 = 2/5

=> x = 2/5 : 11/15 = 6/11

2. a) x(x – 2) + x – 2 = 0

⇔ x(x – 2) + (x – 2) = 0

⇔ (x – 2)(x + 1) = 0

⇔ x – 2 = 0   hoặc   x + 1 = 0

* x – 2 = 0

        x = 2

* x + 1 = 0

      x = -1

Vậy pt có tập nghiệm: S = {2; -1}

b) 5x(x – 3) – x + 3 = 0

⇔ 5x(x – 3) – (x – 3) = 0

⇔ (x – 3)(5x – 1) = 0

⇔ x – 3 = 0   hoặc   5x – 1 = 0

* x – 3 = 0

x = 3

* 5x – 1 = 0

5x = 1

x = 1/5

Vậy pt có tập nghiệm:

S = {3; 1/5}