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

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

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

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

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

\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)

\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

Ta có \(\sqrt{8}+3< \sqrt{9}+3=3+3=6\)

=> \(\sqrt{8}+3< 6\)

Ta có \(\sqrt{48}< \sqrt{49};\sqrt{35}< \sqrt{36}\)

=> \(\sqrt{48}+\sqrt{35}< \sqrt{49}+\sqrt{46}\)

=> \(\sqrt{48}+\sqrt{35}< 13\)

=> \(\sqrt{48}< 13-\sqrt{35}\)

c) Ta có \(-\sqrt{19}< -\sqrt{17}\)

=> \(\sqrt{31}-\sqrt{19}< \sqrt{31}-\sqrt{17}\)

=> \(\sqrt{31}-\sqrt{19}< \sqrt{36}-17=6-\sqrt{17}\)

d) Ta có \(9=\sqrt{81}\Leftrightarrow\sqrt{81}>\sqrt{80}\);

\(-\sqrt{58}>-\sqrt{59}\)

=> \(\sqrt{81}-\sqrt{58}>\sqrt{80}-\sqrt{59}\)

<=> \(9-\sqrt{58}>\sqrt{80}-\sqrt{59}\)

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

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

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

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

Chúc các bạn học tốt và vote cho mình nhé vì đây là lần đầu tiên mình trả lời! Cảm ơn!

$\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

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

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

a)\(\sqrt{8}+3< \sqrt{9}+3=3+3=6< 6+\sqrt{2}\)

b)\(14=\sqrt{196}>\sqrt{195}=\sqrt{13.15}=\sqrt{13}.\sqrt{15}\)

c) Ta có: \(\hept{\begin{cases}\sqrt{27}>\sqrt{25}=5\\\sqrt{6}>\sqrt{4}=2\end{cases}\Rightarrow\sqrt{27}+\sqrt{6}+1>5+2+1=8}\)

Mà \(\sqrt{48}< \sqrt{49}=7< 8\)

\(\Rightarrow\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)

Tham khảo nhé~

\(A=\dfrac{2}{\sqrt{17}+\sqrt{15}}\) ; \(B=\dfrac{2}{\sqrt{15}+\sqrt{13}}\)

Mà \(\sqrt{17}+\sqrt{15}>\sqrt{15}+\sqrt{13}>0\)

\(\Rightarrow\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{15}+\sqrt{13}}\)

\(\Rightarrow A< B\)

\(A=\sqrt{17}-\sqrt{15}=\dfrac{2}{\sqrt{17}+\sqrt{15}}\)

\(B=\sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{13}+\sqrt{15}}\)

mà \(\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{13}+\sqrt{15}}\)

nên A<B

a) <

b) <

c) >

d) <

      a <

            b <

                           c >

                   d <

\(A=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)

\(B=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)

mà \(\sqrt{12}+\sqrt{11}< \sqrt{14}+\sqrt{13}\)

nên A>B