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a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
b/ \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\)
\(=\sqrt{n+1}-1\)
Câu a quy đồng từ từ từ phải qua trái là ra
a) Ta có: \(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}+3\right)\left(\sqrt{2}-\sqrt{3}-3\right)}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-2\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\frac{3\left(\sqrt{2}-\sqrt{3}-3\right)\left(\sqrt{2}+\sqrt{3}\right)}{-\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\frac{3\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}-3\right)}{2}\)
b) Ta có: \(\left(\frac{4}{\sqrt{5}+1}-\frac{4}{\sqrt{5}-1}\right):\sqrt{3+2\sqrt{2}}\)
\(=\left(\frac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\frac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\right):\sqrt{2+2\cdot\sqrt{2}\cdot1+1}\)
\(=\left(\frac{4\left(\sqrt{5}-1\right)}{4}-\frac{4\left(\sqrt{5}+1\right)}{4}\right):\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\left(\sqrt{5}-1-\sqrt{5}-1\right):\left|\sqrt{2}+1\right|\)
\(=-\frac{2}{\sqrt{2}+1}\)(Vì \(\sqrt{2}+1>0\))
\(=-\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)
\(=-2\left(\sqrt{2}-1\right)\)
\(=-2\sqrt{2}+2\)