Bài 1 :
a) \(\frac{4}{\sqrt{5}-\sqrt{3}}-\sqrt{12}=\frac{4\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}-\sqrt{4.3}=\frac{4\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-2\sqrt{3}=2\left(\sqrt{5}+\sqrt{3}\right)-2\sqrt{3}=2\sqrt{5}\)
b) \(\sqrt{\frac{9}{8}}-\sqrt{\frac{49}{2}}+\sqrt{\frac{25}{18}}=\frac{\sqrt{9}}{\sqrt{4.2}}-\frac{\sqrt{49}}{\sqrt{2}}+\frac{\sqrt{25}}{\sqrt{9.2}}\)
\(=\frac{3}{2\sqrt{2}}-\frac{7}{\sqrt{2}}+\frac{5}{3\sqrt{2}}\)
\(=\frac{1}{\sqrt{2}}\left(\frac{3}{2}-7+\frac{5}{3}\right)\)
\(=\frac{1}{\sqrt{2}}.\left(-\frac{23}{6}\right)\)
\(=-\frac{23}{6\sqrt{2}}=-\frac{23\sqrt{2}}{12}\)
Bài 2 :
a) \(\frac{x}{\sqrt{x}-2}=-1\) (ĐKXĐ : \(x\ge0;x\ne4\))
\(\Leftrightarrow x=-\sqrt{x}+2\)
\(\Leftrightarrow x+\sqrt{x}-2=0\)
Đặt \(\sqrt{x}=t\left(t\ge0\right)\)
Ta có : t2 + t - 2 = 0
........ (Tìm t -> thay vào để tìm x -> đối chiếu với đkxđ -> kết luận)
b) \(\sqrt{x-2}=x-4\) (ĐKXĐ : \(x\ge4\))
\(\Leftrightarrow x-2=\left(x-4\right)^2\)
\(\Leftrightarrow x-2=x^2-8x+16\)
\(\Leftrightarrow x^2-8x+16-x+2=0\)
\(\Leftrightarrow x^2-9x+18=0\)
........ (Tìm x -> đối chiếu với đkxđ -> kết luận)
