a.\(x-\sqrt{2x+3}=0\)
\(ĐK:x\ge-\dfrac{3}{2}\)
\(\Leftrightarrow x=\sqrt{2x+3}\)
\(\Leftrightarrow x^2=2x+3\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\) ( vi-ét ) (tm)
b.\(\left\{{}\begin{matrix}\dfrac{x}{a}-y=\dfrac{2}{b}\\x-\dfrac{y}{b}=-\dfrac{1}{a}\end{matrix}\right.\)
\(\left(x;y\right)=\left(3;2\right)\)
\(ĐK:a;b\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{a}-2=\dfrac{2}{b}\\3-\dfrac{2}{b}=-\dfrac{1}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{a}-\dfrac{2}{b}=2\\-\dfrac{1}{a}+\dfrac{2}{b}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}=5\\\dfrac{3}{a}-\dfrac{2}{b}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{2}{5}\\3:\dfrac{2}{5}-\dfrac{2}{b}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{2}{5}\\-\dfrac{2}{b}=-\dfrac{11}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{2}{5}\\b=\dfrac{4}{11}\end{matrix}\right.\)
