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2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
1) Vì \(a,b>0\)\(\Rightarrow\)\(\sqrt{ab}>0\)
\(\Leftrightarrow\)\(2\sqrt{ab}>0\)
\(\Leftrightarrow\)\(a+b+2\sqrt{ab}>a+b\)
\(\Leftrightarrow\)\(\left(\sqrt{a}+\sqrt{b}\right)^2>a+b\)
\(\Leftrightarrow\)\(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
Vậy \(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
1. Ta có: \(\left(\sqrt{a+b}\right)^2=a+b\)
\(\left(\sqrt{a}+\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
Vì \(a>0\), \(b>0\)\(\Rightarrow\sqrt{ab}>0\)\(\Rightarrow2\sqrt{ab}>0\)
\(\Rightarrow a+b< a+2\sqrt{ab}+b\)
\(\Rightarrow\left(\sqrt{a+b}\right)^2< \left(\sqrt{a}+\sqrt{b}\right)^2\)
mà \(\hept{\begin{cases}\sqrt{a+b}>0\\\sqrt{a}+\sqrt{b}>0\end{cases}}\)\(\Rightarrow\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)( đpcm )
\(\dfrac{1}{a}+\dfrac{1}{b}=1\)
\(\Leftrightarrow a+b=ab\)(*)
Xét
\(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\)
\(\Leftrightarrow a+b=a+b-2+2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow2=2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow4=4\left(ab-a-b+1\right)\)
\(\Leftrightarrow4\left(ab-a-b\right)=0\Leftrightarrow ab-a-b=0\)
\(\Leftrightarrow ab=a+b\)(đúng với *)
\(\Rightarrow\)đpcm
Ta có\(\dfrac{1}{a}+\dfrac{1}{b}=1\Leftrightarrow\dfrac{a+b}{ab}=1\Leftrightarrow a+b=ab\Leftrightarrow ab-a-b=0\Leftrightarrow1=ab-a-b+1\Leftrightarrow1=a\left(b-1\right)-\left(b+1\right)\Leftrightarrow1=\left(a-1\right)\left(b-1\right)\Leftrightarrow1=\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow2=2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow0=-2+2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow a+b=a-1+2\sqrt{\left(a-1\right)}\sqrt{\left(b-1\right)}+b-1\Leftrightarrow a+b=\left(\sqrt{a+1}+\sqrt{b+1}\right)^2\Leftrightarrow\sqrt{a+b}=\sqrt{\left(\sqrt{a+1}+\sqrt{b+1}\right)^2}\Leftrightarrow\sqrt{a+b}=\sqrt{a+1}+\sqrt{b+1}\left(đpcm\right)\)
a) \(\sqrt{a}+1>\sqrt{a+1}\)\(\Leftrightarrow\)\(a+2\sqrt{a}+1>a+1\)\(\Leftrightarrow\)\(2\sqrt{a}>0\)( luôn đúng \(\forall x>0\) )
b) \(a-1< a\)\(\Leftrightarrow\)\(\sqrt{a-1}< \sqrt{a}\)
c) \(\left(\sqrt{6}-1\right)^2=6-2\sqrt{6}+1>3-2\sqrt{3.2}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
do \(\sqrt{6}-1>0;\sqrt{3}-\sqrt{2}>0\) nên \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\) ( đpcm )