Question

\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\): \(\frac{1}{\sqrt{a}-\sqrt{b}}\)điều kiện: a>0; b>0; a\(\ne\)b 
a/ Rút gọn B
b/ Tosnh B khi a = \(\sqrt{\left(2+\sqrt{3}\right)}\)và b = \(\sqrt{\left(2-\sqrt{3}\right)}\)

 

Giúp mình với ạ 

Answer 1

\(a)\) \(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=a-b\)

\(b)\) \(B=a-b=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)\(\Rightarrow\)\(B^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)

\(B^2=4-2\sqrt{4-3}=4-2=2\)\(\Rightarrow\)\(B=\sqrt{2}\) ( vì \(B>0\) ) 

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Answer 2

cảm ơn nhe <3 :))