\(\dfrac{1}{\sqrt{2}}P=\dfrac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}}=\dfrac{\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}}{\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}}=\dfrac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2x-1}+1-\sqrt{2x-1}+1}=\dfrac{2\sqrt{x-1}}{2}=\sqrt{x-1}\Rightarrow P=\sqrt{2x-2}\)
+ Đk : \(x\ge1\)
\(P=\sqrt{2}\left(\frac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-\sqrt{2x-1}}}\right)\)
\(P=\sqrt{2}\cdot\frac{\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}}{\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}}\)
\(P=\sqrt{2}\cdot\frac{\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|}{\sqrt{2x-1}+1-\sqrt{2x-1}+1}\)\(=\sqrt{2}\cdot\frac{\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|}{2}\)
\(=\frac{\left(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\right)}{\sqrt{2}}\)
+ Với \(1\le x< 2\) ta có :
\(P=\frac{\sqrt{x-1}+1+1-\sqrt{x-1}}{\sqrt{2}}=\sqrt{2}\)
+ Với \(x\ge2\) ta có :
\(P=\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2}}=\sqrt{2x-2}\)
