a,5(x-1)=x-1 b,(3x-1)^2-(3x-2)^2=0 c,(2x+1)^2-(x-1)^2=0 d,x^2-(y^2-2x+4y+5)=0

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