Đáp án:Mong mod ko xoá :(

 

Giải thích các bước giải:

`a)\sqrt{x^2-9}-\sqrt{4x-12}=0`

`⇔\sqrt{(x-3).(x+3)}-\sqrt{4(x-3)}=0`

`⇔(\sqrt{x-3}).(\sqrt{x+3}-2)=0`

⇔\(\left[ \begin{array}{l}\sqrt{x}=3\\\sqrt{x+3}=2\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=9\\x=1\end{array} \right.\) 

b)`\sqrt{2x^2+8x+8}+\sqrt{x+2}=0`

`\sqrt{(2x+4).(x+2)}+\sqrt{x+2}=0`

`(\sqrt{2x+4}+1).\sqrt{x+2}=0`

\(\left[ \begin{array}{l}\sqrt{2x+4}=-1(vô lí )\\x=-2\end{array} \right.\) 

c)chịu :(

e)`\sqrt{x^2 - 4}-\sqrt{4x + 8}=0`

`\sqrt{(x-2).(x+2)}-2\sqrt{x + 4}=0`

`(\sqrt{(x+2)}-2).\sqrt{x-2}=0`

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

\(\left[ \begin{array}{l}x=14\\x=2\end{array} \right.\) 

d)$\sqrt{2x^2-12x+18}+\sqrt{x-3}=0$ 

$\sqrt{2(x^2-6x+9)}+\sqrt{x-3}=0$ 

$\sqrt{2}.\sqrt{(x-3)^2}+\sqrt{x-3}=0$ 

làm đến đây chiu :(