\(x^2=3+5+2\sqrt{15}=8+\sqrt{60}\)

\(y^2=2+6+2\sqrt{12}=8+\sqrt{48}\)

Mà \(60>48\Rightarrow\sqrt{60}>\sqrt{48}\Rightarrow8+\sqrt{10}>8+\sqrt{48}\)

\(\Rightarrow x^2>y^2\Rightarrow x>y\) (do x;y đều dương)

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

\(Q=\dfrac{1}{\sqrt{2}-1}=\dfrac{\sqrt{2}+1}{2-1}=\sqrt{2}+1\)

Do \(2< \sqrt{2}+1\)

=> P < Q

Ta có: \(12>9\)

\(6\sqrt{3}>4\sqrt{5}\)

Do đó: \(12+6\sqrt{3}>9+4\sqrt{5}\)

\(\Leftrightarrow\sqrt{12+6\sqrt{3}}>\sqrt{9+4\sqrt{5}}\)

Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) (bạn tự c/m) với a = 2003 , b = 2005

được : \(\frac{\sqrt{2003}+\sqrt{2005}}{2}< \sqrt{\frac{2003+2005}{2}}\)

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

Giúp mình với 

\(\sqrt{5-3}=\sqrt{2}\)

\(2>\sqrt{2}\)

\(\Leftrightarrow2>\sqrt{5-3}\)

Bài làm

Ta có: \(28\sqrt{2}\approx39,6\)

           \(\sqrt{14}\approx3,7\)

           \(2\sqrt{147}\approx24,2\)

           \(36\sqrt{4}=72\)

Nên \(36\sqrt{4}>28\sqrt{2}>2\sqrt{147}>\sqrt{14}\left(72>39,6>24,2>3,7\right)\)

Vậy sắp xếp theo thứ tự tăng dần là: \(36\sqrt{4},28\sqrt{2},2\sqrt{147},\sqrt{14}\)

# Học tốt #

\(\sqrt{14}=\sqrt{7}\sqrt{2};2\sqrt{147}=\sqrt{294}\sqrt{2};36\sqrt{4}=\sqrt{2592}\sqrt{2}\)

từ đó so sánh

\(28\sqrt{2}=\sqrt{28^2\cdot2}=\sqrt{784\cdot2}=\sqrt{1568}.\)

\(\sqrt{14}\)

\(2\sqrt{147}=\sqrt{2^2\cdot147}=\sqrt{4\cdot147}=\sqrt{588}\)

\(36\sqrt{4}=\sqrt{36^2\cdot4}=\sqrt{1296\cdot4}=\sqrt{5184}\)

=> Sắp sếp theo thứ tự tăng dần là  \(\sqrt{14};\sqrt{588};\sqrt{1568};\sqrt{5184}\)

=>  \(\sqrt{14};2\sqrt{147};28\sqrt{2};36\sqrt{4}\)

:3

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

\(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)

mà \(\sqrt{2016}+\sqrt{2015}>\sqrt{2014}+\sqrt{2015}\)

nên \(\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

theo ket qua cho thay:9.4594<10

Ta có :

\(\sqrt{3}< \sqrt{4}=2\)

\(\sqrt{8}< \sqrt{9}=3\)

\(\sqrt{24}< \sqrt{25}=5\)

\(\Rightarrow\sqrt{3}+\sqrt{8}+\sqrt{24}< 2+3+5=10\)(đpcm)

Vậy ...

\(\sqrt{3}+\sqrt{8}+\sqrt{24}< \sqrt{4}+\sqrt{9}+\sqrt{25}\)

\(=2+3+5=10\)

Vậy: \(\sqrt{3}+\sqrt{8}+\sqrt{24}< 10\)