A=\(\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}\)
=\(\sqrt{x-2+2\sqrt{x-2}+1}+\sqrt{x-2-2\sqrt{x-2}+1}\)
=\(\sqrt{\left(\sqrt{x-2}+1\right)^2}+\sqrt{\left(\sqrt{x-2}-1\right)^2}\)
=\(\left|\sqrt{x-2}+1\right|+\left|\sqrt{x-2}-1\right|\)
=\(\sqrt{x-2}+1+\left|\sqrt{x-2}-1\right|\)
A=1 <=>\(\sqrt{x-2}+1+\left|\sqrt{x-2}-1\right|\)=1
<=>\(\sqrt{x-2}+\left|\sqrt{x-2}-1\right|\)=0
mà \(\left\{{}\begin{matrix}\sqrt{x-2}\ge0\\\left|\sqrt{x-2}-1\right|\ge0\end{matrix}\right.\)
=>dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x-2}-1=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
=>vô lí
=>pt vô nghiệm
