Đáp án:
Giải thích các bước giải:
a) 3x² + 6x = 0
<=> 3x( x + 2) = 0
Th1 : 3x = 0
<=> x = 0
Th2 : x + 2 = 0
<=> x = - 2
Vậy S = { 0 ; - 2}
b) x³ - 3x² - 4x + 12 = 0
<=> ( x³ - 3x² ) - ( 4x - 12) = 0
<=> x²( x - 3) - 4 ( x - 3) = 0
<=> ( x² - 4)( x - 3) = 0
Th1 : x² - 4 = 0
<=> x = 2 hoặc x = - 2
Th2 : x - 3 = 0
<=> x = 3
Vậy S = { 2 ; -2 ; 3 }
c) ( x - 1)(x + 2) - x - 2 = 0
<=> ( x - 1)( x + 2) - ( x + 2) = 0
<=> ( x + 2)( x - 1 - 1) = 0
<=> ( x + 2)( x - 2) = 0
Th1 : x + 2 = 0
<=> x = - 2
Th2 : x - 2 = 0
<=> x = 2
Vậy S = { - 2 ; 2 }
d) ( 2x + 5)² + ( 4x + 10)(3 - x) = 0
<=> (2x + 5)² + ( 2x + 5)( 6 - 2x) =0
<=> ( 2x + 5)( 2x + 5 + 6 - 2x) = 0
<=> 11 ( 2x + 5) = 0
<=> 2x + 5 = 0
<=> x = - 5/2
Vậy S = { - 5/2 }
e) (3x - 1).(2x + 7) - ( x +1).(6x - 5) = 16
<=> 6x² + 19x - 7 - 6x² - x + 5 = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
Vậy S = { 1 }
g) ( 2x + 3)² - 2( 2x + 3)(2x - 5) +
( 2x - 5)² = x² + 6x + 64
<=> (2x + 3 - 2x + 5)² = x² +6x +64
<=> 8² = x² + 6x + 64
<=> x² + 6x + 64 = 64
<=> x² + 6x = 0
<=> x ( x + 6 ) = 0
Th1 : x = 0
Th2 : x + 6 = 0
<=> x = - 6
Vậy S ={ 0 ; - 6 }
h) ( x - 2)² - ( x +1)( x -1) = 0
<=> x² - 4x + 4 - x² + 1 = 0
<=> - 4x + 5 = 0
<=> - 4x = - 5
<=> x = 5/4
Vậy S = { 5/4 }
i) 3x² - 5x - 12 = 0
<=> ( 3x² - 9x ) + ( 4x - 12 ) = 0
<=> 3x ( x - 3) + 4( x - 3) = 0
<=> ( x - 3)( 3x + 4) = 0
Th1 : x - 3 = 0
<=> x = 3
Th2 : 3x + 4 = 0
<=> 3x = - 4
<=> x = - 4/ 3
Vậy S = { 3 ; -4/3}
k) x² - 4 = 2( x + 2)²
<=> ( x - 2 )( x + 2) - 2( x + 2)² = 0
<=> ( x + 2)[ ( x - 2) - 2( x + 2)] = 0
<=> ( x + 2)( x - 2 - 2x - 4) = 0
<=> ( x + 2) ( - x - 6) = 0
Th1 : x + 2 = 0
<=> x = - 2
Th2 : - x - 6 = 0
<=> - x = 6
<=> x = - 6
Vậy S = { - 2 ; - 6 }
