\(T=\frac{\sqrt{2}.\left(4+\sqrt{7}\right)}{\sqrt{2}.\left(2\sqrt{2}+\sqrt{4+\sqrt{7}}\right)}+\frac{\sqrt{2}.\left(4-\sqrt{7}\right)}{\sqrt{2}.\left(2\sqrt{2}-\sqrt{4-\sqrt{7}}\right)}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{8+2\sqrt{7}}}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{8-2\sqrt{7}}}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{7+2\sqrt{7}+1}}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{7-2\sqrt{7}+1}}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\left(\sqrt{7}+1\right)^2}+\frac{4\sqrt{2}-\sqrt{14}}{4-\left(\sqrt{7}-1\right)^2}\)\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+|\sqrt{7}+1|}+\frac{4\sqrt{2}-\sqrt{14}}{4-|\sqrt{7}-1|}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{7}+1}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{7}+1}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{5+\sqrt{7}}+\frac{4\sqrt{2}-\sqrt{14}}{5-\sqrt{7}}\)

\(T=\frac{\left(4\sqrt{2}+\sqrt{14}\right).\left(5-\sqrt{7}\right)}{\left(5+\sqrt{7}\right).\left(5-\sqrt{7}\right)}+\frac{\left(4\sqrt{2}-\sqrt{14}\right).\left(5+\sqrt{7}\right)}{\left(5+\sqrt{7}\right).\left(5-\sqrt{7}\right)}\)

\(T=\frac{20\sqrt{2}-\sqrt{98}}{9}\)

\(T=\frac{13\sqrt{2}}{9}\)

B = \(\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)

=>  \(\frac{2}{\sqrt{2}}B=\frac{8+2\sqrt{7}}{6+\sqrt{8+2\sqrt{7}}}+\frac{8-2\sqrt{7}}{6-\sqrt{8-2\sqrt{7}}}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{6+\sqrt{7}+1}+\frac{\left(\sqrt{7}-1\right)^2}{6-\sqrt{7}+1}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{\sqrt{7}\left(\sqrt{7}+1\right)}+\frac{\left(\sqrt{7}-1\right)^2}{\sqrt{7}\left(\sqrt{7}-1\right)}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\sqrt{7}+1}{\sqrt{7}}+\frac{\sqrt{7}-1}{\sqrt{7}}=\frac{2\sqrt{7}}{\sqrt{7}}=2\)

=> B = \(\sqrt{2}\)

\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\\ =\dfrac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\right)\\ =\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\right)\\ =\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\right)\\ =\dfrac{1}{\sqrt{2}}\left(\sqrt{7}-1-\sqrt{7}-1\right)\\ =\dfrac{1}{\sqrt{2}}\cdot\left(-2\right)=-\sqrt{2}\)

❤✔✔❤

Đặt biểu thức là A.

Ta có: \(A^2=4-\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4+\sqrt{7}=8-2\sqrt{16-7}=8-6=2\Rightarrow A=\pm2\)

Nhưng do A < 0 nên A = \(-\sqrt{2}\)

\(\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)

\(\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.2+2^2}=\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|=\sqrt{7}-2\)

\(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

\(\sqrt{9+4\sqrt{5}}=\sqrt{5}+2\)

\(\sqrt{11-4\sqrt{7}}=\sqrt{7}-2\)

\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}-\sqrt{2}\)

\(=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7-2\sqrt{7}.1+1}}{\sqrt{2}}-\dfrac{\sqrt{7+2\sqrt{7}.1+1}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)

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

\(=-\sqrt{2}\)

3: \(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

4: \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\sqrt{2}\)

5: \(=\dfrac{\sqrt{23-8\sqrt{7}}}{3}+\dfrac{\sqrt{23+8\sqrt{7}}}{3}\)

\(=\dfrac{4-\sqrt{7}+4+\sqrt{7}}{3}=\dfrac{8}{3}\)

=3.741657387

\(B=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)

\(\Rightarrow B=\dfrac{\sqrt{2}\left(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\right)}{\sqrt{2}}\)

\(\Rightarrow B=\dfrac{\sqrt{2}.\sqrt{4+\sqrt{7}}+\sqrt{2}.\sqrt{4-\sqrt{7}}}{\sqrt{2}}\)

\(\Rightarrow B=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)

\(\Rightarrow B=\dfrac{\sqrt{7+2\sqrt{7}+1}+\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}\)

\(\Rightarrow B=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)

\(\Rightarrow B=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}\)

\(\Rightarrow B=\dfrac{2\sqrt{7}}{\sqrt{2}}\)

\(\Rightarrow B=\sqrt{2}.\sqrt{7}\)

\(\Rightarrow B=\sqrt{14}\)