\(A=3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(=3x+6\sqrt{x}-\left(x-1\right)\)

\(=3x+6\sqrt{x}-x+1\)

\(=2x+6\sqrt{x}+1\)

\(B=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)

\(=x+3\sqrt{x}+\sqrt{x}+3-2\left(x-2\sqrt{x}+1\right)\)

\(=x+4\sqrt{x}+3-2x+4\sqrt{x}-2\)

\(=-x+8\sqrt{x}+1\)

\(C=3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(=3x-3\sqrt{x}-2+\left(\sqrt{x^2}-1\right)\)

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

\(=4x-3\sqrt{x}-3\)

\(D=\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)

\(=x-9-\left(2x-3\sqrt{x}-2\right)\)

\(=x-9-2x+3\sqrt{x}+2\)

\(=-x+3\sqrt{x}-7\)

\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)

\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)

\(=x-4-2\left(2x+3\sqrt{x}-2\right)\)

\(=x-4-4x-6\sqrt{x}+4\)

\(=-3-6\sqrt{x}\)

Ta có:

\(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left[\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right]\\ =\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\left[\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\\ =2:\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

P=\(\left(\dfrac{\sqrt{x^3}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x^3}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\left[\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

P=\(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left[\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}\right]\)

P=\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)

P=\(\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)

P=\(\dfrac{2\sqrt{x}}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$

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

Rút gọn = 0000000000

ĐKXĐ: \(-1\le x\le1\)

Ta có:  \(\sqrt{1+\sqrt{1-x^2}}=\frac{\sqrt{1+x+2\sqrt{1-x^2}+1-x}}{\sqrt{2}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{2}}\)

\(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)\)

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

ĐKXĐ: \(-1\le x\le1\)

Đặt \(a=\sqrt{1-x}>0\)

\(b=\sqrt{1+x}>0\)

\(\Rightarrow a^2+b^2=2\) và \(a^2-b^2=-2x\)

Khi đó: \(B=\frac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\frac{\sqrt{1-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{2-ab}\)

\(=\frac{\sqrt{1-ab}\left(a+b\right)\left(2-ab\right)}{2-ab}\)\(\Rightarrow B=\sqrt{1-ab}\left(a+b\right)\Rightarrow B\sqrt{2}=\sqrt{2-2ab}\left(a+b\right)\)\(=\sqrt{a^2+b^2-2ab}\left(a+b\right)=\left(a-b\right)\left(a+b\right)=a^2-b^2=\)\(-2x\)

\(\Rightarrow b=-\sqrt{2}x\)