Question

So sánh

\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\) và \(\sqrt{3}+1\)

Answer 1

\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(1+\sqrt{12}\right)^2}}}=\sqrt{6+2\sqrt{5-\left|1+\sqrt{12}\right|}}=\sqrt{6+2\sqrt{5-1-\sqrt{12}}}\)

\(=\sqrt{6+2\sqrt{4-\sqrt{12}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2.\left|\sqrt{3}-1\right|}=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)\(=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)

Vậy: \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3}+1\)

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Answer 2

$\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{6+2\sqrt{5-\sqrt{\left(1+\sqrt{12}\right)^2}}}=\sqrt{6+2\sqrt{5-\left|1+\sqrt{12}\right|}=\sqrt{6+2\sqrt{5-1-\sqrt{12}}}=\sqrt{6+2\sqrt{4-\sqrt{12}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2.\left|\sqrt{3}-1\right|}}$$\sqrt{6+2.\left(\sqrt{3}-1\right)}=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(1+\sqrt{3}\right)^2}=\left|1+\sqrt{3}\right|=1+\sqrt{3}$

Vậy √6+2√5−√13+√48 = √3+1

Answer 3

(2√2004)2=4.2004=4008+2.2004(22004)2=4.2004=4008+2.2004

(√2003+√2005)2=2003+2√2003.2005+2005(2003+2005)2=2003+22003.2005+2005

$=4008+2\sqrt{2003.2005}$

So sánh 2004 và $\sqrt{2003.2005}$

Ta có: 

√2003.2005=√(2004−1)(2004+1)=√20042−1<√200422003.2005=(2004−1)(2004+1)=20042−1<20042

Suy ra: 

$\begin{array}{c}2004>\sqrt{2003.2005}\\ \Rightarrow 2.2004>2.\sqrt{2003.2005}\end{array}$

$\Rightarrow 4008+2.2004>4008+2\sqrt{2003.2005}$

⇒(2√2004)2>(√2003+√2005)2⇒(22004)2>(2003+2005)2

Vậy $2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}$