a) \(P=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x\left(1+\sqrt{x}\right)}{x-1+1-\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\cdot\dfrac{x\left(1+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\left(\sqrt{x}+1\right)^2\)
b) thay \(x=\dfrac{2}{2+\sqrt{3}}\) vào P, ta được:
\(P=\left(\sqrt{\dfrac{2}{2+\sqrt{3}}}+1\right)^2=\left(\sqrt{\dfrac{4}{4+2\sqrt{3}}}+1\right)^2=\left(\sqrt{\left(\dfrac{2}{\sqrt{3}+1}\right)^2}+1\right)^2=\left(\dfrac{2}{\sqrt{3}+1}+1\right)^2=\left(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)^2=\left(\sqrt{3}\right)^2=3\)
c) Đk: \(x\ge4\)
\(P\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Rightarrow\left(\sqrt{x}+1\right)^2\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow x\sqrt{x}+2x+\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow x\sqrt{x}+2x-5\sqrt{x}+3=-\sqrt{x-4}\)
tớ nghĩ đề sai vì pt trên vô nghiệm.
