\(\dfrac{1}{\sqrt{48-7}}=\dfrac{1}{\sqrt{41}}=\dfrac{\sqrt{41}}{41}\)

\(\dfrac{1}{\sqrt{48}-\sqrt{7}}=\dfrac{\sqrt{48}+\sqrt{7}}{\left(\sqrt{48}-\sqrt{7}\right)\left(\sqrt{48}+\sqrt{7}\right)}=\dfrac{\sqrt{48}+\sqrt{7}}{41}\)

Ta có

\(\sqrt{41}< \sqrt{48}+\sqrt{7}\Rightarrow\dfrac{1}{\sqrt{48-7}}< \dfrac{1}{\sqrt{48}-\sqrt{7}}\)

b) x và y có thể là số vô tỉ:

Ví dụ: 

\(x=-\sqrt{2};y=\sqrt{2}\Rightarrow x+y=-\sqrt{2}+\sqrt{2}=0\)

\(\dfrac{\Rightarrow x}{y}=\dfrac{-\sqrt{2}}{\sqrt{2}}=-1\)

ngu nhue bò

a/

đk: \(x\ge0\) bình phương 2 vế

\(x^2=x\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b/

dk: \(x\ge0\Rightarrow-\sqrt{x}\le0\)

\(\Rightarrow x=-\sqrt{x}\Rightarrow x=0\)

TL: 

0,(41) = 0,41414141.......

0,4(14) = 0,41414141........

=> 0,(41) = 0,4(14) 

\(\dfrac{1}{2}\widehat{xOy}=\dfrac{2}{3}\widehat{yOz}\Rightarrow\dfrac{2}{4}\widehat{xOy}=\dfrac{2}{3}\widehat{yOz}\Rightarrow\dfrac{1}{4}\widehat{xOy}=\dfrac{1}{3}\widehat{yOz}\)

\(\Rightarrow\dfrac{\widehat{xOy}}{\widehat{yOz}}=\dfrac{4}{3}\)

\(\Rightarrow\widehat{xOy}=\dfrac{180^o}{4+3}.4=\)

\(\Rightarrow\widehat{yOz}=180^o-\widehat{xOy}\)