Giải Phương Trình : x - căn x-2 =4 giúp mik vs ạ , cảm ơn nhiều

`x-\sqrt{x-2}=4`

`ĐKXĐ:x>=2`

Phương trình `<=>x-\sqrt{x-2}-4=0`

`<=>x-2-\sqrt{x-2}+1/4-3/4=0`

`<=>[(x-2)-\sqrt{x-2}+1/4]-3/4=0`

`<=>[(\sqrt{x-2})^2 -2.\sqrt{x-2}. 1/2 +(1/2)^2 ]-(\sqrt{3/4})^2 =0`

`<=>(\sqrt{x-2}-1/2)^2 -((\sqrt{3})/4)^2 =0`

`<=>(\sqrt{x-2}-1/2-(\sqrt{3})/4)(\sqrt{x-2}-1/2+(\sqrt{3})/4)=0`

`<=>(\sqrt{x-2}-(\sqrt{3}+2)/4)(\sqrt{x-2}+(\sqrt{3}-2)/4)=0`

`<=>` \(\left[ \begin{array}{l}\sqrt{x-2}-\dfrac{\sqrt{3}+2}{4}=0\\\sqrt{x-2}+\dfrac{\sqrt{3}-2}{4}=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}\sqrt{x-2}=\dfrac{\sqrt{3}+2}{4}\\\sqrt{x-2}=\dfrac{2-\sqrt{3}}{4}\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x-2=\dfrac{7+4\sqrt{3}}{4}\\x-2=\dfrac{7-4\sqrt{3}}{4}\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac{15+4\sqrt{3}}{4}(tm)\\x=\dfrac{15-4\sqrt{3}}{4}(tm)\end{array} \right.\) 

Vậy `S={(15+4\sqrt{3})/4;(15-4\sqrt{3})/4}`

$\boxed{\bullet\text{ Áp dụng :}\\-(a-b)^2 =a^2 -2ab+b^2\\-(a+b)^2 =a^2 +2ab+b^2\\-a^2 -b^2 =(a-b)(a+b)}$