a, bạn chụp rõ hơn đc ko ?
b, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)đk : \(x\ge3;x\le-3\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
TH1 : x = 3
TH2 : \(\sqrt{x+3}=3\Leftrightarrow x+3=9\Leftrightarrow x=6\)
bổ sung c
\(D=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)
\(\sqrt{2}D=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}=\sqrt{5}-1-\sqrt{5}-1=-2\)
\(D=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)