Lời giải:
\(A=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3+4\sqrt{x-1}}\)
\(=\sqrt{(x-1)-2\sqrt{x-1}+1}+\sqrt{(x-1)+4\sqrt{x-1}+4}\)
\(=\sqrt{(\sqrt{x-1}-1)^2}+\sqrt{(\sqrt{x-1}+2)^2}\)
\(=|\sqrt{x-1}-1|+|\sqrt{x-1}+2|\)
Vì \(x>2\Rightarrow \sqrt{x-1}-1>0\Rightarrow |\sqrt{x-1}-1|=\sqrt{x-1}-1\)
\(\sqrt{x-1}+2>0, \forall 2< x< 5\Rightarrow |\sqrt{x-1}+2|=\sqrt{x-1}+2\)
Do đó:
\(A=(\sqrt{x-1}-1)+(\sqrt{x-1}+2)=2\sqrt{x-1}+1\)
Em thử nhé
\(A=\sqrt{\left(x-1\right)-2\sqrt{x-1}.1+1}+\sqrt{\left(x-1\right)+2.\sqrt{x-1}.2+4}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+2\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+2\right|\)
Do 2 < x < 5 nên \(\sqrt{x-1}-1>\sqrt{2-1}-1=0;\sqrt{x-1}+2>\sqrt{2-1}+2=3>0\)
Do đó \(A=\sqrt{x-1}-1+\sqrt{x-1}+2=2\sqrt{x-1}+1\)
