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Ta có \(16=5+8+3=\sqrt{25}+\sqrt{64}+3.\)
do : \(25>24\Rightarrow\sqrt{25}>\sqrt{24}\); \(64>63\Rightarrow\sqrt{64}>\sqrt{63}\)
=> \(\sqrt{25}+\sqrt{64}+3>\sqrt{24}+\sqrt{63}+3\)
=> \(\sqrt{24}+\sqrt{63}+3< 16\)
ta có căn64>căn63 (1)
căn25>căn24 (2)
167>3 (3)
cộng vế theo vế (1);(2);(3)
=>căn64+căn25+167=16>căn24+căn63+3
\(\frac{\sqrt{10}}{2}=\sqrt{\frac{10}{4}}<\)\(\sqrt{20}=2\sqrt{5}\)
\(\Rightarrow-\frac{\sqrt{10}}{2}>-2\sqrt{5}\)
\(\frac{-\sqrt{10}}{2}=\frac{-\sqrt{2.5}}{2}=\frac{-\sqrt{2}.\sqrt{5}}{2}=-\frac{\sqrt{5}}{\sqrt{2}}=-\sqrt{\frac{5}{2}}>-2\sqrt{5}\)
đúng k
\(a)\) \(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=a-b\)
\(b)\) \(B=a-b=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)\(\Rightarrow\)\(B^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(B^2=4-2\sqrt{4-3}=4-2=2\)\(\Rightarrow\)\(B=\sqrt{2}\) ( vì \(B>0\) )
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Ta có:\(\sqrt{3}-1=-1+\sqrt{3}< -1+\sqrt{4}=-1+2=1\)
\(\Rightarrow\sqrt{3}-1< 1\)
\(\sqrt{3-1}\)=\(-1+\sqrt{3}\)
Vậy \(1>\sqrt{3-1}\)