Coi như các biểu thức đều xác định
a.
\(A=\sqrt{x-1-2\sqrt{x-1}+1}-\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}-\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|-\sqrt{x-1}-1\)
\(\Rightarrow\left[{}\begin{matrix}A=-2\left(\text{nếu }x\ge2\right)\\A=-2\sqrt{x-1}\left(\text{nếu }x\le2\right)\end{matrix}\right.\)
b.
\(B=\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}\)
\(=\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}\)
\(=\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)
\(=\left[{}\begin{matrix}5-2\sqrt{x-1}\left(\text{nếu }x\le5\right)\\1\left(\text{nếu }5\le x\le10\right)\\2\sqrt{x-1}-5\left(\text{nếu }x\ge10\right)\end{matrix}\right.\)
