Câu 3:
b, 27x³ - $\frac{1}{27}$
= (3x)³ - ($\frac{1}{3}$)³
= ( 3x - $\frac{1}{3}$ )[ (3x)² + 3x.$\frac{1}{3}$ + ($\frac{1}{3}$)² ]
= ( 3x - $\frac{1}{3}$ )( 9x² + x + $\frac{1}{9}$ ).
d, x² + 7x + 12
= x² + 3x + 4x + 12
= ( x² + 3x ) + ( 4x + 12 )
= x( x + 3 ) + 4( x + 3 )
= ( x + 3 )( x+ 4 ).
Câu 4:
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 5x = 1
x = $\frac{1}{5}$
Vậy x ∈ { 3; $\frac{1}{5}$ }
