1 SO SÁNH
A, 2005 /2007........... 3/2 B, 131313/717171..........13 72
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2:
=1-1+1-1=0
3:
a: =>34*(100+1)/2:a=17
=>a=101
b: =>5/3(x-1/2)=5/4
=>x-1/2=5/4:5/3=3/4
=>x=5/4
1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)
1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)
Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)
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: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)
=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
\(a,17< 23\Rightarrow333^{17}< 333^{23}\\ b,2007< 2008\Rightarrow2007^{10}< 2008^{10}\\ c,\left(2008-2007\right)^{2009}=1^{2009}=1^{1999}=\left(1998-1997\right)^{1999}\)
\(S=\sqrt[]{1.2007}+\sqrt[]{3.2005}+\sqrt[]{5.2003}+...+\sqrt[]{2007.1}\)
Tổng số hạng của S là :
\(\left(2007-1\right):2+1=1004\left(số,hạng\right)\)
Áp dụng bất đảng Cauchy cho 1004 cặp số \(\left(1;2007\right);\left(3;2005\right);\left(5;2003\right)...\left(2007;1\right)\)
\(\sqrt[]{1.2007}< \dfrac{1+2007}{2}=\dfrac{2008}{2}\)
\(\sqrt[]{3.2005}< \dfrac{3+2005}{2}=\dfrac{2008}{2}\)
\(\sqrt[]{5.2003}< \dfrac{5+2003}{2}=\dfrac{2008}{2}\)
\(.....\)
\(\sqrt[]{2007.1}< \dfrac{2007+1}{2}=\dfrac{2008}{2}\)
\(\Rightarrow S=\sqrt[]{1.2007}+\sqrt[]{3.2005}+\sqrt[]{5.2003}+...+\sqrt[]{2007.1}< 1004.\dfrac{2008}{2}=1004^2\)
Vậy \(S< 1004^2\)
a)
\(\dfrac{-2}{3}\)>\(\dfrac{5}{-8}\)
b)
\(\dfrac{398}{-412}\)<\(\dfrac{-25}{-137}\)
c)
\(\dfrac{-14}{21}\)<\(\dfrac{60}{72}\)
a) 2005/2007 <3/2
b)131313/717171<1372
2005/2007 < 3/2
131313/717171 > 13/72