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\(1)\) Ta có :
\(\left(\sqrt{3\sqrt{2}}\right)^4=\left[\left(\sqrt{3\sqrt{2}}\right)^2\right]^2=\left(3\sqrt{2}\right)^2=9.2=18\)
\(\left(\sqrt{2\sqrt{3}}\right)^4=\left[\left(\sqrt{2\sqrt{3}}\right)^2\right]^2=\left(2\sqrt{3}\right)^2=4.3=12\)
Vì \(18>12\) nên \(\left(\sqrt{3\sqrt{2}}\right)^4>\left(\sqrt{2\sqrt{3}}\right)^4\)
\(\Rightarrow\)\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Vậy \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Chúc bạn học tốt ~
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 )
\(\sqrt{31}+4< \sqrt{36}+4=10\left(1\right)\)
\(6+\sqrt{17}>6+\sqrt{16}=6+4=10\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow\sqrt{31}+4< 10< 6+\sqrt{17}\)
\(\Rightarrow\sqrt{31}+4< \sqrt{17}+6\)
b )
\(\sqrt{3}+\sqrt{2}>\sqrt{1}+\sqrt{1}=2\)
c )
\(\sqrt{12+13}=\sqrt{25}=5\left(1\right)\)
\(\sqrt{12}+\sqrt{13}>\sqrt{4}+\sqrt{9}=2+3=5\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow\sqrt{12+13}< \sqrt{12}+\sqrt{13}\)
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
a,\(\sqrt{12}=2\sqrt{3}=\sqrt{3}+\sqrt{3}\)
ta có \(\sqrt{5}>\sqrt{3}\)và\(\sqrt{7}>\sqrt{3}\)=>\(\sqrt{5}+\sqrt{7}>\sqrt{12}\)
a)A= \(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)=\(\sqrt{6+2\sqrt{3}+2}\)
=> A2=8+2\(\sqrt{3}\)
B=\(\sqrt{3}+1\)=> B2=10+2\(\sqrt{3}\)
=>A>B
Võ Đông Anh Tuấn
Áp dụng \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
a)
\(7=\sqrt{49}\\ 3\sqrt{5}=\sqrt{9}\cdot\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\\ \text{Vì }\sqrt{49}>\sqrt{45}\text{ nên }7>3\sqrt{5}\)
Vậy \(7>3\sqrt{5}\)
b)
\(2\sqrt{7}+3=\sqrt{4}\cdot\sqrt{7}+3=\sqrt{4\cdot7}+3=\sqrt{28}+3\\ \sqrt{28}+3>\sqrt{25}+3=5+3=8\)
Vậy \(8< 2\sqrt{7}+3\)
c)
\(3\sqrt{6}=\sqrt{9}\cdot\sqrt{6}=\sqrt{9\cdot6}=\sqrt{54}\\ 2\sqrt{15}=\sqrt{4}\cdot\sqrt{15}=\sqrt{4\cdot15}=\sqrt{60}\\ \text{Vì } \sqrt{54}< \sqrt{60}\text{nên }3\sqrt{6}< 2\sqrt{15}\)
Vậy \(3\sqrt{6}< 2\sqrt{15}\)
ta có :
\(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{6+2\sqrt{5}}}=\sqrt{\sqrt{\left(1+2\sqrt{5}+5\right)}}\)
\(=\sqrt{\sqrt{\left(1+\sqrt{5}\right)^2}}=\sqrt{1+\sqrt{5}}< \sqrt{1+\sqrt{6}}\)
Vậy \(\sqrt{\sqrt{6+\sqrt{20}}}< \sqrt{1+\sqrt{6}}\)
\(\sqrt{\sqrt{17+12\sqrt{2}}}=\sqrt{\sqrt{9+2.3.2\sqrt{2}+8}}=\sqrt{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\sqrt{3+2\sqrt{2}\sqrt{ }}\)
\(=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(1+\sqrt{2}\right)^2}=\left(1+\sqrt{2}\right)\)
Vậy \(\sqrt{\sqrt{17+12\sqrt{2}}}=1+\sqrt{2}\)