Mình rút gọn như sau:

    \(\left(\sqrt{3-\sqrt{5}}\right).\left(\sqrt{10}-\sqrt{2}\right).\left(3+\sqrt{5}\right)\)

\(=\sqrt{\left(\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}\right)^2}.\left(3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\)

\(=\left(\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}\right).\left(2\sqrt{10}+2\sqrt{2}\right)\)

\(=10+2\sqrt{5}-2\sqrt{5}-2\)

\(=8\)

(Chúc bạn học giỏi và tíck cho mìk vs nhá!)

Câu này em tui đăng chứ đâu biết đâu

IQ vô cực thì tự làm đi

Ta có: A = \(\sqrt{\left(\sqrt{6}-2\sqrt{2}\right)^2}-\sqrt{24-12\sqrt{3}}\)

= \(\left|\sqrt{6}-2\sqrt{2}\right|\) \(-\sqrt{18-2.6\sqrt{3}+6}\)

= \(2\sqrt{2}-\sqrt{6}-\sqrt{\left(\sqrt{18}-\sqrt{6}\right)^2}\)

= \(2\sqrt{2}-\sqrt{6}-\sqrt{18}+\sqrt{6}\)

= \(2\sqrt{2}-3\sqrt{2}=-\sqrt{2}\)

Lời giải:

a)

\(=\frac{(\sqrt{x}+1)\sqrt{x}(\sqrt{x}-\sqrt{y}))\sqrt{x}+\sqrt{y})}{(x-y)x(\sqrt{x}+1)}=\frac{(\sqrt{x}+1)\sqrt{x}(x-y)}{(x-y)x\sqrt{x}+1)}=\frac{1}{\sqrt{x}}\)

b)

\(=\frac{(2-\sqrt{x}-\sqrt{x}-3)(2-\sqrt{x}+\sqrt{x}+3)}{1+2\sqrt{x}}=\frac{(-1-2\sqrt{x}).5}{2\sqrt{x}+1}=\frac{-5(2\sqrt{x}+1)}{2\sqrt{x}+1}=-5\)

\(a,\frac{\left(\sqrt{x}+1\right)\cdot\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\sqrt{x}\left(x+1\right)}\)\(=\frac{\left(\sqrt{x}+1\right)\sqrt{x}\left(x-y\right)}{\left(x-y\right)\sqrt{x} \left(x+1\right)}\)\(=\frac{\sqrt{x}+1}{x+1}\)

\(b,\frac{\left(2-\sqrt{x}\right)^2-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{4+x-4\sqrt{x}-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{1+x-5\sqrt{x}}{1+2\sqrt{x}}\)

a) \(\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}=4^2+\left(\sqrt{2}\right)^2=16+2=18\)

b) \(\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}=9-3=6\)