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\(T=3\sqrt{2}+2\sqrt{3}-\sqrt{3^2.2}+\sqrt{4\left(4-4\sqrt{3}+3\right)}\)
=> \(T=3\sqrt{2}+2\sqrt{3}-3\sqrt{2}+\sqrt{2^2\left(2-\sqrt{3}\right)^2}\)
=> \(T=3\sqrt{2}+2\sqrt{3}-3\sqrt{2}+2\left(2-\sqrt{3}\right)\)
=> \(T=3\sqrt{2}+2\sqrt{3}-3\sqrt{2}+4-2\sqrt{3}\)
=> T=4
2.
A=\(\sqrt{\sqrt{\left(\sqrt{16}-\sqrt{12}\right)^2}}-\sqrt{\sqrt{\left(\sqrt{16}+\sqrt{12}\right)^2}}\)
\(=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{1}\right)^2}\)
\(=\sqrt{3}-1-\left(\sqrt{3}+1\right)\)
\(=\sqrt{3}-1-\sqrt{3}-1\)
\(=-2\)
B= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+\sqrt{1}\right)^2}}}\)
\(=\sqrt{5-2\sqrt{2+\sqrt{8}+1}}\)
\(=\sqrt{5-2\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{5-2\sqrt{\left(\sqrt{2}+\sqrt{1}\right)^2}}\)
\(=\sqrt{5-2\sqrt{2}-2}\)
\(=\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}\)
\(=\sqrt{2}-1\)
a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+\sqrt{14}}{\sqrt{2}\left(\sqrt{6}+\sqrt{14}\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
= \(\sqrt{3.4-3\sqrt{7}}-\sqrt{3.4+3\sqrt{7}}\)
= \(\sqrt{3.\left(4-\sqrt{7}\right)}-\sqrt{3.\left(4+\sqrt{7}\right)}\)
= \(\sqrt{3}.\sqrt{4-\sqrt{7}}-\sqrt{3}.\sqrt{4+\sqrt{7}}\)
= \(\sqrt{3}.\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)\)
\(\)≈ \(-2,449\)
\(\sqrt{\dfrac{13}{4}+\sqrt{3}}-\sqrt{\dfrac{7}{4}-\sqrt{3}}\)
= \(\sqrt{\dfrac{13}{4}+\dfrac{4\sqrt{3}}{4}}-\sqrt{\dfrac{7}{4}-\dfrac{4\sqrt{3}}{4}}\)
= \(\sqrt{\dfrac{13+4\sqrt{3}}{4}}-\sqrt{\dfrac{7-4\sqrt{3}}{4}}\)
= \(\dfrac{\sqrt{13+4\sqrt{3}}}{\sqrt{4}}-\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)
= \(\dfrac{\sqrt{13+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)
≈ \(2,098\)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
b) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) = \(1+\sqrt{2}\)
a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\) = \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\) = \(\dfrac{\sqrt{2}}{2}\)
Ta có: \(\sqrt{28-16\sqrt{3}}=\sqrt{12-2.4.2\sqrt{3}+16}=4-2\sqrt{3}\)
Do đó: \(T=3\sqrt{2}+2\sqrt{3}-3\sqrt{2}+4-2\sqrt{3}=4\)