Bài 4: Giải phương trình a) x(x2-1) = 0 b) ( x -1/2 )( 2x + 5 ) = 0 c/ ( x – 2 ) ( x – 6 ) = 0 d) x22x = 0 e/ (x2 – 2x + 1) – 4 = 0 f/ x(2x – 1) = 0 g/ 4x2 + 4x +1= 0 h) x2 - 5x + 6 = 0 i/ 2x2 + 3x = 0

a) x(x²-1) = 0

TH1: x= 0

TH2: x² -1 = 0

⇔ x² = 1

⇔ x = ±1

Vậy S= {0; ±1}

b) ( x -1/2 )( 2x + 5 ) = 0

TH1: x -1/2 = 0 ⇔ x = 1/2

TH2: 2x +5 = 0 ⇔ x = -5/2

Vậy S= {1/2; -5/2}

c/ ( x – 2 ) ( x – 6 ) = 0

TH1: x-2 = 0 ⇔ x = 2

TH2: x-6 = 0 ⇔ x = 6

Vậy S= {2; 6}

e/ (x² – 2x + 1) – 4 = 0

⇔ (x-1)² -4 = 0

⇔ (x-1 -2)(x-1+2) = 0

⇔ (x-3)(x+1) = 0

TH1: x-3 = 0 ⇔ x = 3

TH2: x+1 = 0 ⇔ x = -1

f/ x(2x – 1) = 0

TH1: x= 0

TH2: 2x -1 = 0 ⇔ x = 1/2

Vậy S= {0; 1/2}

g/ 4x² + 4x +1= 0

⇔ (2x +1)² = 0

⇔ 2x +1 = 0

⇔ x = -1/2

Vậy S= {-1/2}

h) x² - 5x + 6 = 0

⇔ x² -2x -3x +6 = 0

⇔ x(x-2) -3(x -2) = 0

⇔ (x-2)(x-3) = 0

TH1: x-2 = 0 ⇔ x = 2

TH2: x-3 = 0 ⇔ x= 3

Vậy S= {2;3}

i/ 2x² + 3x = 0

⇔ x(2x +3) = 0

TH1: x= 0

TH2: 2x +3 = 0 ⇔ x = -3/2

Vậy S= {-3/2}