Question

2. rút gọn biểu thức

P=\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

Answer 1

P = \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)DKXD: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

   = \(\sqrt{x}+\sqrt{x}\)

   = \(2\sqrt{x}\)

Vậy tại x ∈ ĐKXĐ thì P = \(2\sqrt{x}\)

Answer 2

Ta có: \(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

\(=\sqrt{x}+\sqrt{x}\)

\(=2\sqrt{x}\)