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\(8^5=\left(2^3\right)^5=2^{15}=2.2^{14}\)
\(3.4^7=3.\left(2^2\right)^7=3.2^{14}\)
Vì 2 < 3 nên 85 < 3 . 47
\(\dfrac{5}{2}=\sqrt{\dfrac{25}{4}}>\sqrt{\dfrac{20}{4}}=\sqrt{5}\)
=>\(\sqrt{5}< \dfrac{5}{2}\)
=>căn 5-5/2<0
`5/2 = 2,5 = sqrt(6,25)`
Do `5 < 6,25` nên `sqrt 5 - 5/2 < 0`.
Giả sử \(\sqrt[]{2}+\sqrt[]{3}< 3\)
\(\Leftrightarrow\left(\sqrt[]{2}+\sqrt[]{3}\right)^2< 3^2\)
\(\Leftrightarrow2+3+2\sqrt[]{6}< 9\)
\(\Leftrightarrow2\sqrt[]{6}< 4\)
\(\Leftrightarrow\left(2\sqrt[]{6}\right)^2< 4^2\)
\(\Leftrightarrow24< 16\left(sai\right)\)
Vậy \(\sqrt[]{2}+\sqrt[]{3}>3\)
Ta có :
\(\sqrt{2}+\sqrt{3}=\left(\sqrt{2}+\sqrt{3}\right)^2\\ =\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2\\ =2+2\sqrt{6}+3\\ =5+2\sqrt{6}\left(5+2\sqrt{6}>5+2\sqrt{4}\right)\)
Mà \(5+2\sqrt{4}=5+2\cdot2=5+4=9\)
\(\Rightarrow5+2\sqrt{6}>9\)
Nên \(\sqrt{2}+\sqrt{3}>3\)
16 = 9 + 7 = 9 + √49
9 + 4√5 = 9 + √80
Do 49 < 80 nên √49 < √80
⇒ 9 + √49 < 9 + √80
Vậy 16 < 9 + 4√5
Ta có:
\(16=9+5=9+\sqrt{25}\)
\(9+4\sqrt{5}=9+\sqrt{4^2\cdot5}=9+\sqrt{20}\)
Mà: \(20< 25\)
\(\Rightarrow\sqrt{20}< \sqrt{25}\)
\(\Rightarrow9+\sqrt{20}< 9+\sqrt{25}\)
\(\Rightarrow9+4\sqrt{5}< 16\)
(2-căn 2)^2=6-4*căn 2=1/4+23/4-4căn 2
(1/2)^2=1/4
mà 23/4-4căn 2>0
nên 2-căn 2>1/2
Ta có: \(\left\{{}\begin{matrix}\sqrt{2008}+\sqrt{2005}< \sqrt{2015}+\sqrt{2009}\\\sqrt{2010}+\sqrt{2007}< \sqrt{2015}+\sqrt{2009}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{\sqrt{2008}+\sqrt{2005}}+\dfrac{1}{\sqrt{2010}+\sqrt{2007}}>\dfrac{2}{\sqrt{2015}+\sqrt{2009}}\)
\(\Leftrightarrow\dfrac{\sqrt{2008}-\sqrt{2005}}{3}+\dfrac{\sqrt{2010}-\sqrt{2007}}{3}>\dfrac{\sqrt{2015}-\sqrt{2009}}{3}\)
\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}\)