CMR \(\sqrt{2\sqrt{3}\sqrt{4...\sqrt{ }}2003}< 3\)
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H9
HT.Phong (9A5)
CTVHS
29 tháng 8 2023
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 \(x=\dfrac{3\sqrt{2}}{2}\)
...
29 tháng 8 2023
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: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)
=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
4 tháng 9 2018
a) \(A=\sqrt{a-2-2\sqrt{a-3}}-\sqrt{a+1-4\sqrt{a-3}}=\sqrt{\left(a-3\right)-2\sqrt{a-3}+1}-\sqrt{\left(a-3\right)-4\sqrt{a-3}+4}=\sqrt{\left(\sqrt{a-3}-1\right)^2}-\sqrt{\left(\sqrt{a-3}-2\right)^2}\)Ta có 3≤a≤4\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\left(\sqrt{a-3}-1\right)^2}=1-\sqrt{a-3}\\\sqrt{\left(\sqrt{a-3}-2\right)^2}=2-\sqrt{a-3}\end{matrix}\right.\)
Vậy A=\(1-\sqrt{a-3}-\left(2-\sqrt{a-3}\right)=1-\sqrt{a-3}-2+\sqrt{a-3}=-1\)b) B=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\times\sqrt{2003-2\sqrt{2005-2\sqrt{2004}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+9}}}\times\sqrt{2003-2\sqrt{2004-2\sqrt{2004}+1}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\times\sqrt{2003-2\sqrt{\left(\sqrt{2004}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\times\sqrt{2003-2\sqrt{2004}+2}=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\times\sqrt{2004-2\sqrt{2004}+1}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\times\sqrt{\left(\sqrt{2004}-1\right)^2}=\sqrt{\sqrt{5}-\sqrt{5}+1}\times\left(\sqrt{2004}-1\right)=\sqrt{1}\times\left(\sqrt{2004}-1\right)=\sqrt{2004}-1\)
ND
8 tháng 7 2018
\(\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}+\sqrt{8}+\sqrt{10}}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)+\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{4}+\sqrt{2}.\sqrt{5}}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)\left(1+\sqrt{2}\right)}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)
\(=1+\sqrt{2}\)
⇒ ĐPCM
HT
2
14 tháng 8 2017
Ta có:
\(\sqrt{2\sqrt{3\sqrt{4....\sqrt{2017}}}}\)
< \(\sqrt{2\sqrt{3\sqrt{4...\sqrt{2016\sqrt{2018}}}}}\)
\(=\sqrt{2\sqrt{3\sqrt{4...\sqrt{2017^2-1}}}}\)
< \(\sqrt{2\sqrt{3\sqrt{4...\sqrt{2015.2017}}}}\)
.......................................................................
< \(\sqrt{2.4}< \sqrt{9}=3\)
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0