\(\dfrac{\sqrt{3}+\sqrt{a}}{\sqrt{5}+\sqrt{b}}\in Q\)
\(\Leftrightarrow\dfrac{\sqrt{3}+\sqrt{a}}{\sqrt{5}+\sqrt{b}}=\dfrac{m}{n}\left(m;n\in Z^+;\left(m;n\right)=1\right)\)
\(\Leftrightarrow n\left(\sqrt{3}+\sqrt{a}\right)=m\left(\sqrt{5}+\sqrt{b}\right)\)
\(\Leftrightarrow n\sqrt{3}-m\sqrt{5}+n\sqrt{a}-m\sqrt{b}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}n\sqrt{3}-m\sqrt{5}=0\\n\sqrt{a}-m\sqrt{b}=0\end{matrix}\right.\) \(\left(a;b\in Z^+\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=3\end{matrix}\right.\) thỏa mãn yêu cầu đề bài.
Sửa lại
a√3-m√b+n√a-m√5=0
a√3-m√b=0
n√a-m√5=0
→a=5;b=3;m=n=1
