a: Ta có: \(E=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)

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

\(=\left(\dfrac{4\sqrt{x}+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}}{x-1}\)

\(=\dfrac{4x^2}{\left(x-1\right)^2}\)

b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-2x^2+4x-2=0\)

\(\Leftrightarrow2x^2+4x-2=0\)

\(\Leftrightarrow x^2+2x-1=0\)

\(\Leftrightarrow\left(x+1\right)^2=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{2}-1\\x=\sqrt{2}-1\end{matrix}\right.\)

c: Ta có: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\)

Thay x=2 vào E, ta được:

\(E=\dfrac{4\cdot2^2}{1}=16\)

a/ \(B=(\dfrac{2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-6}{x-9}):(1+\dfrac{6}{x-9})\)

   =  \((\dfrac{2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-9}{x-9}+\dfrac{6}{x-9})\)

   =\((\dfrac{2(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-3}{x-9})\)

   =\((\dfrac{2\sqrt{x}+6+\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-3}{x-9})\)

   =\((\dfrac{2\sqrt{x}+6+\sqrt{x}-6}{x-9}).(\dfrac{x-9}{x-3})\)

    = \(\dfrac{3\sqrt{x}}{x-3}\)

Vậy B=\(\dfrac{3\sqrt{x}}{x-3}\)

b/ Để B≥0 thì \(\dfrac{3\sqrt{x}}{x-3} \)≥0

                   \(<=>\begin{cases} x-3 không= 0\\ 3\sqrt{x}>/0 \end{cases} \)

                     <=> \(\begin{cases} x không= 3\\ x>/0 \end{cases} \)

Vậy để B≥0 thì x không = 3 và x≥0

a) Ta có: \(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(1+\dfrac{2}{\sqrt{x}-1}\right)\)

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

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

b) Để C<-1 thì C+1<0

\(\Leftrightarrow\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}}< 0\)

\(\Leftrightarrow2\sqrt{x}-1< 0\)

\(\Leftrightarrow x< \dfrac{1}{4}\)

Kết hợp ĐKXĐ, ta được: \(0< x< \dfrac{1}{4}\)

1. ĐKXĐ: $x>0; x\neq 9$

\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)

2. ĐKXĐ: $x\geq 0; x\neq 4$

\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)

\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)

3. ĐKXĐ: $a\geq 0; a\neq 1$

\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)

\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

a: Sửa đề: \(C=\left(\dfrac{2x+1}{\sqrt{x^3}+1}-\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\right)\left(\dfrac{1-\sqrt{x^3}}{1+\sqrt{x^3}}-\sqrt{x}\right)\)

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

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

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

Lời giải:
a. ĐKXĐ: $a\geq 0; a\neq 1$

b.

\(P=\left[\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}+1\right].\left[\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}-1\right].\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}\)

\(=(\sqrt{a}+1)(\sqrt{a}-1).\sqrt{2}=\sqrt{2}(a-1)\)

c.

\(P=\sqrt{2}(\sqrt{2+\sqrt{2}}-1)=\sqrt{4+2\sqrt{2}}-\sqrt{2}\)

a. ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ne1\\\sqrt{a}\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

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

\(=\left[\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right].\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right].\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\cdot\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\cdot\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)

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

\(=\sqrt{2}a-\sqrt{2}\)