Đáp án:

Bai 4 :

$a. ( x - 2 )( x + y )$

$b. ( x + z )( y - 2 )$

$c. ( x - y )( x - 1 )$

$d. ( y + 1 )( x + 1 )$

$e. x( x - 2 )^{2}$

$g. ( x - y - 3 )( x - y + 3 )$

$k. ( x - y )( x - y + 4 )$

Bai 5 :

$a.$ \(\left[ \begin{array}{l}x=0\\x=±2\end{array} \right.\) 

$b.$ \(\left[ \begin{array}{l}x=1\\x=-\frac{1}{3}\end{array} \right.\) 

$c.$ \(\left[ \begin{array}{l}x=\frac{3}{5}\\x=-\frac{3}{5}\end{array} \right.\) 

$d. x = \frac{4}{9}$

$e.$ \(\left[ \begin{array}{l}x=0\\x=±3\end{array} \right.\) 

$g.$ \(\left[ \begin{array}{l}x=-5\\x=\frac{2}{3}\end{array} \right.\) 

Bai 6 :

$a. ( x - 1 )( x - 2 )$

$b. ( x + 3 )( x - 2 )$

$c. x( 2x - 3 )^{2}$

$d. ( x - 2y - 3z )( x - 2y + 3z )$

$g. ( x - 4 )( x - 3 )$

Giải thích các bước giải:

Áp dụng hằng đẳng thức đáng nhớ sau vào bài : 

+) $( a + b )^{2} = a^{2} + 2ab + b^{2}$

+) $( a - b )^{2} = a^{2} - 2ab + b^{2}$

+) $a^{2} - b^{2} = ( a - b )( a + b )$

+) $a^{3} + b^{3} = ( a + b )( a^{2} - ab + b^{2} )$

Bài 4 :

$a. x^{2} - 2x + xy - 2y$

$= x( x - 2 ) + y( x - 2 )$

$= ( x - 2 )( x + y )$

$b. xy + yz - 2( x + z )$

$= y( x + z ) - 2( x + z )$

$= ( x + z )( y - 2 )$

$c. x^{2} - xy - x + y$

$= x( x - y ) - ( x - y )$

$= ( x - y )( x - 1 )$

$d. xy + 1 + x + y$

$= ( xy + x ) + ( y + 1 )$

$= x( y + 1 ) + ( y + 1 )$

$= ( y + 1 )( x + 1 )$

$e. x^{3} - 4x^{2} + 4x$

$= x( x^{2} - 4x + 4 )$

$= x( x - 2 )^{2}$

$g. x^{2} - 2yx + y^{2} - 9$

$= ( x - y )^{2} - 3^{2}$

$= ( x - y - 3 )( x - y + 3 )$

$k. x^{2} + y^{2} - 2xy + 4x - 4y$

$= ( x - y )^{2} + 4( x - y )$

$= ( x - y )( x - y + 4 )$

Bai 5 :

$a. x^{3} - 4x = 0$

⇔ $x( x^{2} - 4 ) = 0$

⇔ \(\left[ \begin{array}{l}x=0\\x^{2}=4\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=0\\x=±2\end{array} \right.\) 

$b. 4x^{2} - ( x + 1 )^{2} = 0$

⇔ $( 2x - x - 1 )( 2x + x + 1 ) = 0$

⇔ $( x - 1 )( 3x + 1 ) = 0$

⇔ \(\left[ \begin{array}{l}x=1\\3x=-1\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=1\\x=-\frac{1}{3}\end{array} \right.\) 

$c. 25x^{2} - 9 = 0$

⇔ $( 5x - 3 )( 5x + 3 ) = 0$

⇔ \(\left[ \begin{array}{l}5x=3\\5x=-3\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{3}{5}\\x=-\frac{3}{5}\end{array} \right.\) 

$d. ( x + 2 )( x^{2} - 2x + 4 ) - x( x + 3 )( x - 3 ) = 12$

⇔ $x^{3} + 2^{3} - x( x^{2} - 3^{2} ) = 12$

⇔ $x^{3} + 8 - x^{3} + 9x = 12$

⇔ $9x + 8 = 12$

⇔ $9x = 12 - 8$

⇔ $9x = 4$

⇔ $x = \frac{4}{9}$

$e. x^{3} - 9x = 0$

⇔ $x( x^{2} - 9 ) = 0$

⇔ \(\left[ \begin{array}{l}x=0\\x^{2}=9\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=0\\x=±3\end{array} \right.\) 

$g. 3x( x + 5 ) - 2x - 10 = 0$

⇔ $3x( x + 5 ) - 2( x + 5 ) = 0$

⇔ $( x + 5 )( 3x - 2 ) = 0$

⇔ \(\left[ \begin{array}{l}x=-5\\3x=2\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-5\\x=\frac{2}{3}\end{array} \right.\) 

Bai 6 :

$a. x^{2} - 3x + 2$

$= ( x^{2} - x ) - ( 2x - 2 )$

$= x( x - 1 ) - 2( x - 1 )$

$= ( x - 1 )( x - 2 )$

$b. x^{2} + x - 6$

$= ( x^{2} + 3x ) - ( 2x + 6 )$

$= x( x + 3 ) - 2( x + 3 )$

$= ( x + 3 )( x - 2 )$

$c. 4x^{3} - 12x^{2} + 9x$

$= x( 4x^{2} - 12x + 9 )$

$= x( 2x - 3 )^{2}$

$d. x^{2} - 4xy + 4y^{2} - 9z^{2}$

$= ( x - 2y )^{2} - (3z)^{2}$

$= ( x - 2y - 3z )( x - 2y + 3z )$

$g. x^{2} - 7x + 12$

$= ( x^{2} - 4x ) - ( 3x - 12 )$

$= x( x - 4 ) - 3( x - 4 )$

$= ( x - 4 )( x - 3 )$