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\(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{5+2\sqrt{5}+1}}=\sqrt{\sqrt{\left(\sqrt{5}+1\right)^2}}=\sqrt{\sqrt{5}+1}< \sqrt{\sqrt{6}+1}\)
\(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{6+2\sqrt{5}}}=\sqrt{\sqrt{\left(\sqrt{5}+1\right)^2}}=\sqrt{\sqrt{5}+1}\)
Vì \(\sqrt{\sqrt{5}+1}< \sqrt{\sqrt{6}+1}\Rightarrow\sqrt{\sqrt{6+\sqrt{20}}}< \sqrt{1+\sqrt{6}}\)
a ) VT = \(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{6+\sqrt{4.5}}}=\sqrt{\sqrt{6+2\sqrt{5}}}=\sqrt{\sqrt{\left(1+\sqrt{5}\right)^2}}=\sqrt{1+\sqrt{5}}\)
Có 5 < 6 => \(\sqrt{5}< \sqrt{6}\Rightarrow\sqrt{1+\sqrt{5}}< \sqrt{1+\sqrt{6}}\)
Vậy \(\sqrt{\sqrt{6+\sqrt{20}}}< \sqrt{1+\sqrt{6}}\)
b) VT = \(\sqrt{\sqrt{17+12\sqrt{2}}}=\sqrt{\sqrt{17+2.2\sqrt{2}.3}}=\sqrt{\sqrt{\left(2\sqrt{2}+3\right)^2}=\sqrt{2\sqrt{2}+3}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
=> VT = VP
=> \(\sqrt{\sqrt{17+12\sqrt{2}}}=\sqrt{2}+1\)
c) \(\sqrt{\sqrt{28-16\sqrt{3}}}=\sqrt{\sqrt{16-2.4.2\sqrt{3}+12}}=\sqrt{\sqrt{\left(4-2\sqrt{3}\right)^2}}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
Có -1 > -2 => \(\sqrt{3}-1>\sqrt{3}-2\Rightarrow\sqrt{\sqrt{28-16\sqrt{3}}}>\sqrt{3}-2\)
a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)
mà 112<117
nên \(4\sqrt{7}< 3\sqrt{13}\)
b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
mà \(\dfrac{21}{4}>\dfrac{36}{7}\)
nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)
d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)