\(=2+1+2\sqrt{2}=\left(\sqrt{2}\right)^2+1+2\sqrt{2}=\left(1+\sqrt{2}\right)^2\)

\(A=\dfrac{\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{3}\cdot3\sqrt{2}\)

\(=\sqrt{2}\left(\sqrt{2}-1\right)+\sqrt{2}=2\)

Bạn cần làm gì với biểu thức này thì bạn ghi rõ ra.

Lời giải:
ĐKXĐ: $x>0; x\neq 1$

\(P=\frac{1}{\sqrt{x}+1}+\frac{x}{\sqrt{x}(1-\sqrt{x})}=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\)

\(=\frac{1-\sqrt{x}+\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(1-\sqrt{x})}=\frac{x+1}{1-x}\)

b. Khi $x=\frac{1}{\sqrt{2}}$ thì:

\(P=\frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}\)

C=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{\left(\sqrt{x}+2\right).\left(x-1\right)-\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{x\sqrt{x}-\sqrt{x}+2x-2-\left(x-1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{x-1+x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{\left(x-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}\)

a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)

=3căn 6-6-3căn 6=-6

b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)

\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)

\(1-2x\ge0\Leftrightarrow2x\le1\Leftrightarrow x\le\frac{1}{2}\)

\(x-\sqrt{1-2x}>0\)

X>0