Bài 1:
a: ĐKXĐ: \(x\le\dfrac{5}{2}\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)
Bài 1:
a) ĐKXĐ: \(x\le\dfrac{5}{2}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}1+x\ge0\\x-2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)
Bài 3:
a) ĐKXĐ: \(x\ge-2\)
\(pt\Leftrightarrow\sqrt{x+2}+\sqrt{x+2}=4\sqrt{x+2}-6\)
\(\Leftrightarrow2\sqrt{x+2}=6\Leftrightarrow\sqrt{x+2}=3\)
\(\Leftrightarrow x+2=9\Leftrightarrow x=7\left(tm\right)\)
b) ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
c) ĐKXĐ: \(x\ge1\)
\(pt\Leftrightarrow x^2+2=x^2-2x+1\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(ktm\right)\)
Vậy \(S=\varnothing\)
d) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-1\Leftrightarrow\left|x-3\right|=2x-1\)
\(\left[{}\begin{matrix}x-3=2x-1\left(x\ge3\right)\\x-3=1-2x\left(\dfrac{1}{2}\le x< 3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)
