d, ĐK: \(x\ge3\)
\(\sqrt{x-3}-2\sqrt{x^2-9}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{11}{4}\left(l\right)\end{matrix}\right.\)
e, ĐK: \(x\ge3\)
\(2\sqrt{x+3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow2\sqrt{x+3}+\sqrt{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(2+\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\sqrt{x+3}=0\)
\(\Leftrightarrow x=-3\left(l\right)\)
\(\Rightarrow\) Phương trình vô nghiệm.
\(d,\sqrt{x-3}-2\sqrt{x^2-9}=0\left(x\ge3\right)\\ \Leftrightarrow\sqrt{x-3}-2\sqrt{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x+3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x-3}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\\sqrt{x-3}=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(N\right)\\x=\dfrac{13}{4}\left(N\right)\end{matrix}\right.\)
\(e,2\sqrt{x+3}+\sqrt{x^2-9}=0\left(x\ge3\right)\\ \Leftrightarrow\sqrt{x+3}\left(2+\sqrt{x-3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\left(L\right)\\x\in\varnothing\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
À câu e làm như này cho đơn giản này.
ĐK: \(x\ge3\)
\(2\sqrt{x+3}+\sqrt{x^2-9}\ge2\sqrt{3+3}+\sqrt{3^2-9}=2\sqrt{6}>0\)
\(\Rightarrow\) Phương trình vô nghiệm.
