giải pt sau căn x^2-x-2=căn x+1

`ĐKXĐ : x \ge 2` ; x ≤ -1

`\sqrt{x^2 - x - 2} = \sqrt{x+1}`

`<=> \sqrt{x^2 + x - 2x - 2} - \sqrt{x+1} = 0``

`<=> \sqrt{(x+1)(x-2)} - \sqrt{x+1} = 0`

`<=> \sqrt{x+1}(\sqrt{x-2} - 1) = 0`

`<=>` \(\left[ \begin{array}{l}\sqrt{x+1}=0\\\sqrt{x-2} - 1=0\end{array} \right.\) 

`TH1 : \sqrt{x+1}=0`

`<=> x+1 = 0 <=> x = -1 (tm)`

`TH2 : \sqrt{x-2} - 1=0`

`<=> \sqrt{x-2} = 1`

`<=> x-2 = 1`

`<=> x = 3 (tmđk)`

Vậy pt có `2` nghiệm `x = 3` `; x = -1`