a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2-6x+9=x+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2-7x+6=0\end{matrix}\right.\Leftrightarrow x=6\)
e: \(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2-4=x^2-4x+4\end{matrix}\right.\Leftrightarrow x=2\)
f: \(\Leftrightarrow\sqrt{2x+3}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(x-3\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow x=3\)
`a)`\(\sqrt{x+3}=x-3\); \(ĐK:x\ge3\)
\(\Leftrightarrow x+3=x^2-6x+9\)
\(\Leftrightarrow x^2-7x+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{6\right\}\)
`b)`\(x+2\sqrt{x+1}=2\) ;\(ĐK:x\ge-1\)
\(\Leftrightarrow x+1+2\sqrt{x+1}+1=4\)
\(\Leftrightarrow\left(\sqrt{x+1}+1\right)^2=4\)
\(\Leftrightarrow\sqrt{x+1}+1=2\) ( vì \(\sqrt{x+1}+1>0\) )
\(\Leftrightarrow x=0\left(tm\right)\)
Vậy \(S=\left\{0\right\}\)
`c)`\(\sqrt{x^2+9}-\sqrt{x^2-x-3}=2\)
\(ĐK:x^2-x-3\ge0\)
\(\Leftrightarrow\sqrt{x^2+9}=2+\sqrt{x^2-x-3}\)
\(\Leftrightarrow x^2+9=4+x^2-x-3+4\sqrt{x^2-x-3}\)
\(\Leftrightarrow8+x=4\sqrt{x^2-x-3}\)
\(\Leftrightarrow64+16x+x^2=16\left(x^2-x-3\right)\)
\(\Leftrightarrow64+16x+x^2=16x^2-16x-48\)
\(\Leftrightarrow15x^2-32x-112=0\)
\(\Leftrightarrow\left(x-4\right)\left(15x+28\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-\dfrac{28}{15}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{4;-\dfrac{28}{15}\right\}\)
`d)`\(\sqrt{x+1}-\sqrt{x-7}=\sqrt{12-x}\)
\(ĐK:7\le x\le12\)
\(\Leftrightarrow x+1+x-7-2\sqrt{\left(x+1\right)\left(x-7\right)}=12-x\)
\(\Leftrightarrow3x-18-2\sqrt{\left(x+1\right)\left(x-7\right)}=0\)
\(\Leftrightarrow3\left(x-6\right)=2\sqrt{\left(x+1\right)\left(x-7\right)}\)
\(\Leftrightarrow9\left(x^2-12x+36\right)=4\left(x+1\right)\left(x-7\right)\)
\(\Leftrightarrow9x^2-108x+324=4\left(x^2-6x-7\right)\)
\(\Leftrightarrow9x^2-108x+324=4x^2-24x-28\)
\(\Leftrightarrow5x^2-84x+352=0\)
\(\Leftrightarrow\left(x-8\right)\left(5x-44\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=\dfrac{44}{5}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{8;\dfrac{44}{5}\right\}\)
`e)`\(\sqrt{x^2-4}=x-2\)
\(ĐK:x\ge2\)
\(\Leftrightarrow x^2-4=x^2-4x+4\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow x=2\left(tm\right)\)
Vậy \(S=\left\{2\right\}\)
`f)`\(x-\sqrt{2x+3}=0\)
\(ĐK:x\ge-\dfrac{3}{2}\)
\(\Leftrightarrow x=\sqrt{2x+3}\)
\(\Leftrightarrow x^2=2x+3\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1;3\right\}\)
