a)<=>2x+5=1-x
=>x=-4
b)<=>x^2=3
=>x=\(\sqrt{3}\)
a) \(\sqrt{2x+5}=\sqrt{1-x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+5\ge0\\1-x\ge0\\2x+5=1-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{2}\\x\le1\\3x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{5}{2}\le x\le1\\x=-\dfrac{4}{3}\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
- Vậy \(S=\left\{-\dfrac{4}{3}\right\}\)
b) \(\sqrt{x^2-x}=\sqrt{3-x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x\ge0\\3-x\ge0\\x^2-x=3-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-1\right)\ge0\left(1\right)\\x\le3\\x^2-3=0\end{matrix}\right.\)
- Giải bất phương trình (1):
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-1\ge0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x\le0\\x-1\le0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le0\end{matrix}\right.\)
- Phương trình đã cho tương đương:
\(\left\{{}\begin{matrix}x\ge1hayx\le0\\x\le3\\\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le3hayx\le0\\x=\sqrt{3}\left(nhận\right)hayx=-\sqrt{3}\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
- Vậy \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)
