Giải phương trình sau : 1. 3x - $\dfrac{1}{x-2 }$= $\dfrac{x-1 }{2-x }$ 2. $\dfrac{2}{x^2-4 }$ - $\dfrac{1}{x(x-2 ) }$ + $\dfrac{x-4 }{x(x+2) }$=0

Đáp án:

Giải thích các bước giải:

 `1;3x-1/[x-2]=[x-1]/[2-x](dk:x\ne2)`

`<=>[3x^2 -6x-1]/[x-2]=[1-x]/[x-2]`

`<=>3x^2 -6x-1=1-x`

`<=>3x^2 -6x+x-1-1=0`

`<=>3x^2 -5x-2=0`

`<=>3x^2 -6x +x-2=0`

`<=>3x(x-2)+(x-2)=0`

`<=>(x-2)(3x+1)=0`

`<=>`\(\left[ \begin{array}{l}x-2=0\\3x+1=0\end{array} \right.\) $\\$`<=>`\(\left[ \begin{array}{l}x=2(l)\\x=\dfrac{-1}{3}\end{array} \right.\) $\\$Vậy `S={-1/3 }`

`2;2/[x^2 -4]-1/[x(x-2)]+[x-4]/[x(x+2)]=0(đk:x\ne0;-2;2)`

`<=>[2x-x-2+(x-4)(x-2)]/[x(x-2)(x+2)]=0`

`<=>x-2+x^2 -4x-2x+8=0`

`<=>x^2 -5x+6=0`

`<=>(x-2)(x-3)=0`

`<=>`\(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\) $\\$`<=>`\(\left[ \begin{array}{l}x=2(l)\\x=3\end{array} \right.\) $\\$Vậy `S={3}`