b) ( 2x + 3)2 – (2x + 1)(2x – 1) = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
Vậy x = 1
c) (4x + 3)(4x – 3) – (4x - 5)2 = 16
<=> 16x2 - 9 - 16x2 + 40x - 25 = 16
<=> 40x - 34 = 16
<=> 40x = 50
<=> \(\frac{5}{4}\)
Vậy \(x=\frac{5}{4}\)
e) (x + 1)3 – x2(x + 3) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 2
<=> x = \(\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
f) (x – 2)3 – x(x – 1)(x + 1) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x3 + x + 6x2 = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
Vậy x = 1