a: ĐKXĐ: \(\left\{{}\begin{matrix}2x+1>=0\\x-1>=0\end{matrix}\right.\)
=>x>=1
\(\sqrt{2x+1}=\sqrt{x-1}\)
=>2x+1=x-1
=>2x-x=-1-1
=>x=-2(loại)
b: ĐKXĐ: \(x^2+x>=0\)
=>x(x+1)>=0
=>\(\left[{}\begin{matrix}x>=0\\x+1< =0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>=0\\x< =-1\end{matrix}\right.\)
\(\sqrt{x^2+x}=x\)
=>\(\left\{{}\begin{matrix}x^2+x=x^2\\x>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x>=0\end{matrix}\right.\)
=>x=0(nhận)
c: ĐKXĐ: \(\left\{{}\begin{matrix}1-x^2>=0\\x-1>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2< =1\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1< =x< =1\\x>=1\end{matrix}\right.\)
=>x=1
Khi x=1 thì \(\sqrt{1-x^2}=\sqrt{1-1^2}=0;\sqrt{x-1}=\sqrt{1-1}=0\)
Do đó: \(\sqrt{1-x^2}=\sqrt{x-1}\) khi x=1
=>x=1 là nghiệm của phương trình
d:
ĐKXĐ: \(x^2-1>=0\)
=>\(x^2>=1\)
=>\(\left[{}\begin{matrix}x>=1\\x< =-1\end{matrix}\right.\)
\(\sqrt{x^2-1}-x+1=0\)
=>\(\sqrt{x^2-1}=x-1\)
=>\(\left\{{}\begin{matrix}x-1>=0\\\left(x-1\right)^2=x^2-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2-2x+1-x^2+1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=1\\-2x+2=0\end{matrix}\right.\Leftrightarrow x=1\)(nhận)
e: ĐKXĐ: \(x^2-4>=0\)
=>\(x^2>=4\)
=>\(\left[{}\begin{matrix}x>=2\\x< =-2\end{matrix}\right.\)
\(\sqrt{x^2-4}-x+2=0\)
=>\(\sqrt{x^2-4}=x-2\)
=>\(\left\{{}\begin{matrix}x-2>=0\\x^2-4=\left(x-2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2-4-x^2+4x-4=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=2\\4x-8=0\end{matrix}\right.\)
=>x=2(nhận)
f:
\(\sqrt{x^2-3}=\sqrt{4x-3}\)
=>\(\left\{{}\begin{matrix}4x-3>=0\\x^2-3=4x-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{4}\\x^2-4x=0\end{matrix}\right.\Leftrightarrow x=4\)
g: ĐKXĐ: \(1-2x^2>=0\)
=>\(2x^2< =1\)
=>\(x^2< =\dfrac{1}{2}\)
=>\(-\dfrac{\sqrt{2}}{2}< =x< =\dfrac{\sqrt{2}}{2}\)
\(\sqrt{1-2x^2}=x-1\)
=>\(\left\{{}\begin{matrix}x-1>=0\\1-2x^2=\left(x-1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\-2x^2+1=x^2-2x+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-1\\-3x^2+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\x\left(3x-2\right)=0\end{matrix}\right.\)
=>\(x\in\left\{0;\dfrac{2}{3}\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;\dfrac{2}{3}\right\}\)
