Đáp án:
Bài 3 :
$a.$ \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
$b.$ \(\left[ \begin{array}{l}x=0\\x=±2\end{array} \right.\)
$c.$ \(\left[ \begin{array}{l}x=0\\x=\frac{5}{3}\end{array} \right.\)
$d.$ \(\left[ \begin{array}{l}x=3\\x=\frac{1}{2}\end{array} \right.\)
$e. x = 1$
$f. x = \frac{-1}{2}$
$g.$ \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\)
$h. x = 1$
Giải thích các bước giải:
Bài 3 :
$a. 2x^{2} - 6x = 0$
⇔ $2x( x - 3 ) = 0$
⇔ \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
$b. x^{3} - 4x = 0$
⇔ $x( x^{2} - 4 ) = 0$
⇔ \(\left[ \begin{array}{l}x=0\\x^{2}=4\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=±2\end{array} \right.\)
$c. 10x - 6x^{2} = 0$
⇔ $2x( 5 - 3x ) = 0$
⇔ \(\left[ \begin{array}{l}x=0\\x=\frac{5}{3}\end{array} \right.\)
$d. 2x( x - 3 ) - ( x - 3 ) = 0$
⇔ $( x - 3 )( 2x - 1 ) = 0$
⇔ \(\left[ \begin{array}{l}x=3\\x=\frac{1}{2}\end{array} \right.\)
$e. x^{2} - 2x + 1 = 0$
⇔ $x( x - 1 ) - ( x - 1 ) = 0$
⇔ $( x - 1 )( x - 1 ) = 0$
⇔ $x = 1$
$f. 4x^{2} + 4x = -1$
⇔ $2x( 2x + 1 ) + ( 2x + 1 ) = 0$
⇔ $( 2x + 1 )( 2x + 1 ) = 0$
⇔ $x = \frac{-1}{2}$
$g. ( x + 1 )( x - 3 ) - ( x + 1 ) = 0$
⇔ $( x + 1 )( x - 3 -1 ) = 0$
⇔ $( x + 1 )( x - 4 ) = 0$
⇔ \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\)
$h. x^{3} - 3x^{2} + 3x - 1 = 0$
⇔ $x^{2}( x - 1 ) - 2x( x - 1 ) + ( x - 1 ) = 0$
⇔ $( x - 1 )( x^{2} - 2x + 1 ) = 0$
⇔ $( x - 1 )[ x( x - 1 ) - ( x - 1 ) ] = 0$
⇔ $( x - 1 )( x - 1 )( x - 1 ) = 0$
⇔ $x = 1$
Bài 4 :
$9x( x - y ) - 10( y - x )^{2} = 0$
⇔ $9x( x - y ) - 10( x - y )^{2} = 0$
⇔ $( x - y )[ 9x - 10( x - y ) ] = 0$
⇔ $( x - y )( 9x - 10x + 10y ) = 0$
⇔ $( x - y )( 10y - x ) = 0$
⇔ $10y = x$ ( do $x \ne y$ )
Bài 5 :
$a. 27^{5} - 3^{11} = (3^{3})^{5} - 3^{11}$
$= 3^{15} - 3^{11} = 3^{11}( 3^{4} - 1 )$
$= 3^{11}( 81 - 1 ) = 3^{11}×80$
⇒ $( 27^{5} - 3^{11} )$ chia hết cho $80$
$b. 39^{20} + 39^{13} = 39^{13}( 39^{7} + 1 )$
$39^{7} + 1 = 39^{6}(39+1)-39^{5}(39+1)+39^{4}(39+1)-39^{3}(39+1)+39^{2}(39+1)-39(39+1)+(39+1)$
⇔ $39^{7} + 1 = (39+1)(39^{6}-39^{5}+39^{4}-39^{3}+39^{2}-39+1)$
⇔ $39^{7} + 1 = 40(39^{6}-39^{5}+39^{4}-39^{3}+39^{2}-39+1)$
⇒ $39^{20} + 39^{13} = 39^{13}×40×(39^{6}-39^{5}+39^{4}-39^{3}+39^{2}-39+1)$
⇒ $39^{20} + 39^{13}$ chia hết cho $40$
