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

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

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

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

Với `a > 0,a \ne 1` có:

`C=(\sqrt{a}/[\sqrt{a}-1]-\sqrt{a}/[a-\sqrt{a}]):[\sqrt{a}+1]/[a-1]`

`C=[a-\sqrt{a}]/[\sqrt{a}(\sqrt{a}-1)].[(\sqrt{a}-1)(\sqrt{a}+1)]/[\sqrt{a}+1]`

`C=[\sqrt{a}(\sqrt{a}-1)]/[\sqrt{a}(\sqrt{a}-1)].[\sqrt{a}-1)(\sqrt{a}+1)]/[\sqrt{a}+1]`

`C=\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 

có 4 hàng hàng số 2 mấy bạn kéo giùm mình cái phần dấu gạch chia ở dưới dài ra để kéo dài cả hai biểu thức luôn được ko dấu gạch dưới phần căn a-1 với căn a+1 đó ạ mình ko biết kéo dài ra rồi các bạn làm bình thường giúp mình nha mình đang rất cần làm ơn

ĐK: `a \ne 1 ; a>=0` 

`C=(\sqrta/(\sqrta-1)-\sqrta/(a-1)) : (\sqrta+1)/(a-1)`

`=(\sqrta/(\sqrta-1) - \sqrta/((\sqrta-1)(\sqrta+1))) . (a-1)/(\sqrta+1)`

`= (\sqrta(\sqrta+1)-\sqrta)/((\sqrta-1)(\sqrta+1)) . ((\sqrta-1)(\sqrta+1))/(\sqrta+1)`

`= a/(\sqrta+1)`

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

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

\(=\dfrac{a}{\sqrt{a}+1}\)

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

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)

a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(đk:a>0,x\ne1\right)\)

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

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

\(=\dfrac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)

b) \(A=-2\sqrt{a}>-6\)

\(\Leftrightarrow\sqrt{a}< 3\Leftrightarrow0\le a< 9\) và \(a\ne1\)

c) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)

\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)

\(a,A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\left(dk:x\ge0,x\ne1\right)\)

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

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

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

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

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

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

\(b,x=14-6\sqrt{5}=\sqrt{5^2}-2.3.\sqrt{5}+3^2=\left(\sqrt{5}-3\right)^2\)

\(\Rightarrow A=\dfrac{\sqrt{\left(\sqrt{5}-3\right)^2}}{14-6\sqrt{5}+\sqrt{\left(\sqrt{5}-3\right)^2}+1}\)

\(=\dfrac{\left|\sqrt{5}-3\right|}{-6\sqrt{5}+15+\left|\sqrt{5}-3\right|}\)

\(=\dfrac{3-\sqrt{5}}{-6\sqrt{5}+15+3-\sqrt{5}}\)

\(=\dfrac{3-\sqrt{5}}{18-7\sqrt{5}}\)

\(c,A=1\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=1\Leftrightarrow\sqrt{x}-x-\sqrt{x}-1=0\Leftrightarrow-x-1=0\Leftrightarrow x=-1\left(ktm\right)\)

Vậy khi A = 1 thì không có giá trị x thỏa mãn.

a) \(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

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

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

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

\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

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

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

\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)

\(A=\dfrac{a-1}{\sqrt{a}}\)

b) Ta có:

\(a=4+2\sqrt{3}=\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2=\left(\sqrt{3}+1\right)^2\)

Thay vào A ta có:

\(A=\dfrac{\left(\sqrt{3}+1\right)^2-1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{4+2\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{3+2\sqrt{3}}{\sqrt{3}+1}\) 

c) \(A< 0\) khi:

\(\dfrac{a-1}{\sqrt{a}}< 0\)

Mà: \(\sqrt{a}\ge0\forall x\) (xác định) 

\(\Leftrightarrow a-1< 0\)

\(\Leftrightarrow a< 1\)

Kết hợp với đk:

\(0< a< 1\)