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\(\frac{B}{\sqrt{2}}=\frac{\frac{2+\sqrt{3}}{2}}{\sqrt{2}+\sqrt{\frac{4+2\sqrt{3}}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\sqrt{2}-\sqrt{\frac{4-2\sqrt{3}}{2}}}\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}+\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}-\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}}\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}-\frac{\sqrt{3}-1}{\sqrt{2}}}=\frac{\frac{2+\sqrt{3}}{2}}{\frac{\sqrt{3}+3}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{3-\sqrt{3}}{\sqrt{2}}}\)
\(=\frac{\left(2+\sqrt{3}\right).\sqrt{2}}{2\cdot\left(3+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right).\sqrt{2}}{2.\left(3-\sqrt{3}\right)}\)
=> \(B=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(B=\frac{3+\sqrt{3}}{6}+\frac{3-\sqrt{3}}{6}=1\)
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Vài chỗ mình làm vắn tắt không hiểu cứ hỏi nhé, còn kết quả mình ấn máy tính ra chính xác rùi :)
Ta có \(P=\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)
\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{2\left(1-\sqrt{5}\right)}\right).\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}\right).\left(\sqrt{7}-\sqrt{3}\right)=\frac{\sqrt{7}+\sqrt{3}}{2}.\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\frac{7-3}{2}=2\)
Vậy \(P=2\)
\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)
\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)
\(\Leftrightarrow C=-3\)
Ta có: \(\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{2\sqrt{3}}{3}+\frac{\sqrt{2}}{3}+\frac{2\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{12}.\sqrt{\frac{1}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{12\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{5-2\sqrt{6}}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left(\sqrt{3}-\sqrt{2}\right)\)(vì \(\sqrt{3}-\sqrt{2}>0\))
\(=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{3}=\sqrt{3}\)