các cao nhân giải giúp em bài toán này với ạ! Tìm x, biết: a, x^2 - 3x = 0 b, ( x + 2 )^2 - ( x - 2) ( x + 2) = 0 c, x^2 - 4 + ( x - 2 )^2 = 0 d, (2x - 1 )^2 - ( x + 3 )^2 = 0 e, x^2 + 5x - 4 = 0

`a, x^2 - 3x = 0`

`⇔x(x-3)=0`

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

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

             Vậy `S={0; 3}`

`b, ( x + 2 )^2 - ( x - 2) ( x + 2) = 0`

`⇔(x+2)[x+2-(x-2)]=0`

`⇔(x+2)(x+2-x+2)=0`

`⇔4(x+2)=0`

`⇔x+2=0`

`⇔x=-2`

                   Vậy `S={-2}`

`c, x^2 - 4 + ( x - 2 )^2 = 0`

`⇔(x-2)(x+2)+(x-2)^2=0`

`⇔(x-2)(x+2+x-2)=0`

`⇔2x(x-2)=0`

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

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

                        Vậy `S={0; 2}`

`d, (2x - 1 )^2 - ( x + 3 )^2 = 0`

`⇔[2x-1-(x+3)](2x-1+x+3)=0`

`⇔(2x-1-x-3)(3x+2)=0`

`⇔(x-4)(3x+2)=0`

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

`⇔`\(\left[ \begin{array}{l}x=4\\x=-2/3\end{array} \right.\) 

                      Vậy `S={4; -2/3}`