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những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
1, \(=\frac{4\sqrt{2}\left(2-\sqrt{2}\right)}{2^2-\sqrt{2}^2}-\frac{4\sqrt{2}\left(2+\sqrt{2}\right)}{2^2-\sqrt{2}^2}\)
=\(\frac{4\sqrt{2}\left(2-\sqrt{2}\right)}{2}-\frac{4\sqrt{2}\left(2+\sqrt{2}\right)}{2}\)
=\(2\sqrt{2}\left(2-\sqrt{2}\right)-2\sqrt{2}\left(2+\sqrt{2}\right)\)
=\(4\sqrt{2}-4-4\sqrt{2}-4\)
=-8
2, =\(\sqrt{2}+\sqrt{2}-2.3\sqrt{2}+\left|1-\sqrt{2}\right|\)
= \(-4\sqrt{2}+1-\sqrt{2}\) = \(1-5\sqrt{2}\)
3, =\(9\sqrt{\frac{2.2}{3.2}}+5\sqrt{9.6}-\sqrt{\frac{1}{6}}\)
=\(3\sqrt{6}+15\sqrt{6}-\frac{1}{6}\sqrt{6}\)
=\(\frac{107}{6}\sqrt{6}\)
4, =\(\sqrt{\left(4+2\sqrt{2}\right)\left(4-2\sqrt{2}\right)}.\left(2\sqrt{2}-\sqrt{2}\right)\)
= \(\sqrt{4^2-\left(2\sqrt{2}\right)^2}.\sqrt{2}\)
= \(\sqrt{16-8}.\sqrt{2}\)
= \(\sqrt{8}.\sqrt{2}=\sqrt{16}=4\)
5, = \(\sqrt{9-2.3.\sqrt{5}+5}+\sqrt{1-2.1.\sqrt{2}+2}+\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}\)
\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{(1-\sqrt{2})^2}+\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)\(=\left|3-\sqrt{5}\right|+\left|1-\sqrt{2}\right|+\left|\sqrt{5}-\sqrt{2}\right|\)
\(=3-\sqrt{5}+1-\sqrt{2}+\sqrt{5}-\sqrt{2}\)
\(=4-2\sqrt{2}\)
h)
\(H=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2-\sqrt{3})(2+\sqrt{3})}}=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=2\sqrt{3}\)
i)
\(I=\frac{2+\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3.1}}}+\frac{2-\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3.1}}}=\frac{2+\sqrt{3}}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{(\sqrt{3}-1)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{6}{6}=1\)
ê)
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+1+2(\sqrt{2}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+1+2(\sqrt{2}+\sqrt{5})}=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}=\sqrt{2}+\sqrt{5}+1\)
g)
\(13+\sqrt{48}=13+2\sqrt{12}=12+1+2\sqrt{12.1}=(\sqrt{12}+1)^2\)
\(\Rightarrow \sqrt{13+\sqrt{48}}=\sqrt{12}+1\)
\(\Rightarrow \sqrt{3+\sqrt{13+\sqrt{48}}}=\sqrt{4+\sqrt{12}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
\(\Rightarrow 2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}=2\sqrt{2-\sqrt{3}}=\sqrt{2}.\sqrt{4-2\sqrt{3}}=\sqrt{2}.\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{2}(\sqrt{3}-1)=\sqrt{6}-\sqrt{2}\)
\(\Rightarrow G=1\)