\(3\sqrt{3}-2\sqrt{2}=\sqrt{27}-\sqrt{8}>\sqrt{25}-\sqrt{9}=5-3\)

\(\Rightarrow3\sqrt{3}-2\sqrt{2}>2\)

c.

(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})

Mà:

\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)

\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)

\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)

Lời giải:

a.

$5+\sqrt{2}>5+\sqrt{1}=6$

$4+\sqrt{3}< 4+\sqrt{4}=6$

$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$

b.

$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$

$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$

Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$

2) \(-x^2+4x-2\)

\(=-\left(x^2-4x+2\right)\)

\(=-\left(x^2-4x+4-2\right)\)

\(=-\left(x-2\right)^2+2\)

Ta có: \(-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+2\le2\forall x\)

Dấu "=" xảy ra:

\(\Leftrightarrow-\left(x-2\right)^2+2=2\Leftrightarrow x=2\)

Vậy: GTLN của bt là 2 tại x=2

b) \(\sqrt{2x^2-3}\) (ĐK: \(x\ge\sqrt{\dfrac{3}{2}}\))

Mà: \(\sqrt{2x^2-3}\ge0\forall x\)

Dấu "=" xảy ra:

\(\sqrt{2x^2-3}=0\Leftrightarrow x=\sqrt{\dfrac{3}{2}}=\dfrac{3\sqrt{2}}{2}\)

Vậy GTNN của bt là 0 tại \(x=\dfrac{3\sqrt{2}}{2}\)

...

1:

b: \(4\sqrt{5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{75}\)

=>\(4\sqrt{5}>5\sqrt{3}\)

=>\(\sqrt{4\sqrt{5}}>\sqrt{5\sqrt{3}}\)

c: \(3-2\sqrt{5}-1+\sqrt{5}=2-\sqrt{5}< 0\)

=>\(3-2\sqrt{5}< 1-\sqrt{5}\)

d: \(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

\(\sqrt{2005}-\sqrt{2004}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

\(\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)

=>\(\dfrac{1}{\sqrt{2006}+\sqrt{2005}}< \dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

=>\(\sqrt{2006}-\sqrt{2005}< \sqrt{2005}-\sqrt{2004}\)

e: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)

\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)

=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)

=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)

\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)

\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

Do đó: A=B

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

\(\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}\right)^3+1^3+3.2+3\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

--> Bằng nhau

a)

Có: 

\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)

Vì \(\sqrt{117}>\sqrt{116}\)  nên \(3\sqrt{13}>2\sqrt{29}\)

b)

Có:

\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)

\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)

Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\)  nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)

c)

Có:

\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)

\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)

Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)

d)

Có:

\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)

\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)

lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)

 \(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)

3⁵⁴=(3²)²⁷=9²⁷

2⁸¹=(2³)²⁷=8²⁷

Vì 9²⁷>8²⁷ nên 3⁵⁴>2⁸¹

Ta có: 3⁵⁴ = (3²)²⁷=9²⁷

Và: 2⁸¹=(2³)²⁷=8²⁷

Vì 9>8 nên 9²⁷>8²⁷

Vậy 3⁵⁴ >2⁸¹

so sánh3^54 va 2^81

3⁵⁴=(3²)²⁷=9²⁷

2⁸¹=(2³)²⁷=8²⁷

Vì 9²⁷>8²⁷ nên 3⁵⁴>2⁸¹

nhé !

\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)

\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

Đặt: 

\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

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

\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\right)\)

\(A=\dfrac{1}{\sqrt{2}}\left(\left|1+\sqrt{5}\right|+\left|\sqrt{5}-1\right|\right)\)

\(A=\dfrac{1}{\sqrt{2}}\left(1+\sqrt{5}+\sqrt{5}-1\right)\)

\(A=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

Ta có: \(A^2=\left(\sqrt{10}\right)^2=10\)  

\(B=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

Mà: \(4\sqrt{5}>1\)

Nên: \(A^2< B^2\)

\(\Rightarrow A< B\)

Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

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

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}+1+\sqrt{5}-1\right)=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

=>A^2=(căn 10)^2=10=9+1

Đặt B=2+căn 5

=>B^2=(2+căn 5)^2=9+4căn 5

1<4căn 5

=>9+1<9+4căn 5

=>A^2<B^2

=>A<B

Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(\Rightarrow A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=6+2\sqrt{9-5}=6+2.2=10\)

\(B=2+\sqrt{5}\Rightarrow B^2=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

\(>9+1=10=A^2\)

\(\Rightarrow B^2>A^2\Rightarrow B>A\)

Vậy, B>A

\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

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