Đáp án:
Giải thích các bước giải:
a ) 5x( x - 1 ) = x - 1
→ 5x( x - 1 ) - ( x - 1 ) = 0
→ ( x - 1 )( 5x - 1 ) = 0
→ \(\left[ \begin{array}{l}x-1=0\\5x-1=0\end{array} \right.\)
→ \(\left[ \begin{array}{l}x=1\\x=\frac{1}{5}\end{array} \right.\)
b ) $(3x-1)^{2}$ - $(3x-2)^{2}$ = 0
→ ( 3x - 1 + 3x - 2 )( 3x - 1 - 3x + 2 ) = 0
→ ( 6x - 3 ) . 1 = 0
→ 6x - 3 = 0
→ 6x = 3
→ x = $\frac{1}{2}$
c ) $(2x+1)^{2}$ - $(x-1)^{2}$ = 0
→ ( 2x + 1 + x + 1 )( 2x + 1 - x + 1 ) = 0
→ 3x( x + 2 ) = 0
→ \(\left[ \begin{array}{l}3x=0\\x+2=0\end{array} \right.\)
→ \(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\)
d ) $x^{2}$ + $y^{2}$ - 2x + 4y + 5 = 0
→ ( $x^{2}$ - 2x + 1 )( $y^{2}$ + 4y + 4 ) = 0
→ $(x-1)^{2}$ + $(y+2)^{2}$ = 0
Vì $(x-1)^{2}$ ≥ 0
$(y+2)^{2}$ ≥ 0
Mà $(x-1)^{2}$ + $(y+2)^{2}$ = 0
⇒ $(x-1)^{2}$ = 0
→ x = 1
⇒ $(y+2)^{2}$ = 0
→ y = -2
