a: \(\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\)

\(\left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

mà \(-2\sqrt{105}>-2\sqrt{120}\)

nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

b: \(\left(\sqrt{2}+\sqrt{8}\right)^2=10+2\cdot4=16=12+4\)

\(\left(3+\sqrt{3}\right)^2=12+6\sqrt{3}\)

mà \(4< 6\sqrt{3}\)

nên \(\sqrt{2}+\sqrt{8}< 3+\sqrt{3}\)

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\(a,\left(\sqrt{\sqrt{3}}\right)^4=3< 4=\left(\sqrt{2}\right)^4\Rightarrow\sqrt{\sqrt{3}}< \sqrt{2}\\ b,\left(\sqrt{2\sqrt{3}}\right)^4=12< 18=\left(\sqrt{3\sqrt{2}}\right)^4\Rightarrow\sqrt{2\sqrt{3}}=\sqrt{3\sqrt{2}}\\ c,\left(2+\sqrt{6}\right)^2=8+4\sqrt{6};5^2=25=8+17;\left(4\sqrt{6}\right)^2=96< 289=17^2\\ \Rightarrow4\sqrt{6}< 17\Rightarrow2+\sqrt{6}< 5\\ d,\left(7-2\sqrt{2}\right)^2=57-28\sqrt{2};4^2=16=57-41;\left(28\sqrt{2}\right)^2=1568< 41^2=1681\\ \Rightarrow28\sqrt{2}< 41\Rightarrow7-2\sqrt{2}>4\\ e,\left(\sqrt{15}+\sqrt{8}\right)^2=23+4\sqrt{30};7^2=49=23+26;\left(4\sqrt{30}\right)^2=240< 676=26^2\\ \Rightarrow4\sqrt{30}< 26\Rightarrow\sqrt{15}+\sqrt{8}< 7\)

\(f,\left(\sqrt{37}-\sqrt{14}\right)^2=51-2\sqrt{518};\left(6-\sqrt{15}\right)^2=51-12\sqrt{15};\left(2\sqrt{518}\right)^2=2072;\left(12\sqrt{15}\right)^2=2160\\ \Rightarrow2\sqrt{518}< 12\sqrt{15}\Rightarrow\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)

\(A=\sqrt{12+\sqrt{12+\sqrt{12}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}< \sqrt{12+\sqrt{12+\sqrt{16}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}\)\(=7\)

\(B=\sqrt{14}+\sqrt{11}>\sqrt{13,69}+\sqrt{10,89}=7\)

\(\Rightarrow A< B\)

Ta có:

 \(12< 16\Rightarrow\sqrt{12}< \sqrt{16}=4\\ 6< 9\Rightarrow\sqrt{6}< \sqrt{9}=3\)

\(\Rightarrow A< \sqrt{12+\sqrt{12+4}}+\sqrt{6+\sqrt{6+\sqrt{6+3}}}=\sqrt{12+4}+\sqrt{6+3}=4+3=7\) (1)

Lại có :

\(B=\sqrt{14}+\sqrt{11}\Rightarrow B^2=25+2\sqrt{14.11}=25+2\sqrt{154}>25+2\sqrt{144}=25+2.12=49=7^2\)

Mà B > 0

\(\Rightarrow B>7\) (2)

Từ (1),(2) suy ra A<B

Ta có \(6=\sqrt{36}\)

\(\sqrt{37}-\sqrt{14}=\sqrt{37}+\left(-\sqrt{14}\right)\)

\(6-\sqrt{15}=\sqrt{36}-\sqrt{15}=\sqrt{36}+\left(-\sqrt{15}\right)\)

Vì \(\sqrt{37}>\sqrt{36}\) và \(-\sqrt{14}>-\sqrt{15}\)

\(\Rightarrow\sqrt{37}+\left(-\sqrt{14}\right)>\sqrt{36}+\left(-\sqrt{15}\right)\)

\(\Rightarrow\sqrt{37}-\sqrt{14}>\sqrt{36}-\sqrt{15}\)

hay \(\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)

Chúc bn học tốt

Đặt A = \(\sqrt{15}\)-\(\sqrt{14}\)và B = \(\sqrt{14}\)-\(\sqrt{13}\)(A, B >0)

A^2 = 29-2\(\sqrt{15.14}\) và B^2 = 27 -2\(\sqrt{14.13}\)

A^2-B^2 = 2-2(\(\sqrt{15.14}\)+\(\sqrt{14.13}\)) <0

=> A^2 < B^2 => A<B

a ) \(\sqrt{37}\) và \(6\)

Ta có : \(6=\sqrt{36}\)

mà \(\sqrt{36}< \sqrt{37}\)

\(\Rightarrow\sqrt{37}>6\)

b ) \(2\sqrt{3}\) và \(3\sqrt{2}\)

Ta có : \(2\sqrt{3}=\sqrt{12}\)

\(3\sqrt{2}=\sqrt{18}\)

mà : \(\sqrt{12}< \sqrt{18}\)

\(\Rightarrow2\sqrt{3}< 3\sqrt{2}\)

c ) \(\sqrt{63}+\sqrt{35}\) và \(14\)

Ta có : \(\sqrt{63}< \sqrt{64}=8\) và \(\sqrt{35}< \sqrt{36}=6\)

\(\Rightarrow\sqrt{63}+\sqrt{35}< 8+6=14\)