a) \(M=\left(\dfrac{2x+3\sqrt{x}}{x\sqrt{x}+1}+\dfrac{1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\left(x>0\right)\)

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

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

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

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

b) Ta có: \(\sqrt{x}+4>\sqrt{x}+1\Rightarrow\dfrac{\sqrt{x}+4}{\sqrt{x}+1}>1\)

c) \(\dfrac{\sqrt{x}+4}{\sqrt{x}+1}=1+\dfrac{3}{\sqrt{x}+1}\)

Ta có: \(\left\{{}\begin{matrix}3>0\\\sqrt{x}+1>0\end{matrix}\right.\Rightarrow1+\dfrac{3}{\sqrt{x}+1}>1\Rightarrow M>1\)

Lại có: \(\sqrt{x}+1>1\left(x>0\right)\Rightarrow\dfrac{3}{\sqrt{x}+1}< 3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}< 4\Rightarrow M< 4\)

\(\Rightarrow1< M< 4\Rightarrow M\in\left\{2;3\right\}\)

\(M=2\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=2\Rightarrow\dfrac{3}{\sqrt{x}+1}=1\Rightarrow\sqrt{x}+1=3\)

\(\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(M=3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=3\Rightarrow\dfrac{3}{\sqrt{x}+1}=2\Rightarrow2\sqrt{x}+2=3\)

\(\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

Bài 1:

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

=2

Bài 2:

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

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

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

Đặt \(\sqrt{x}=a\)

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

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

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

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

\(\Rightarrow B=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Vậy...

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

\(a.R=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}\right)\)

\(R=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+xy-1}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]:\left[\dfrac{xy-1-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]\)

\(R=\dfrac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}+xy-1}{xy-1}:\dfrac{xy-1-x\sqrt{y}+\sqrt{x}+\sqrt{xy}+1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}}{xy-1}\)

\(R=\dfrac{-2\sqrt{x}-2}{xy-1}:\dfrac{-2x\sqrt{y}-2\sqrt{xy}}{xy-1}\)

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

\(R=\dfrac{\sqrt{x}+1}{x\sqrt{y}+\sqrt{xy}}\)

\(b.C=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{7\sqrt{x}+4}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(C=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{7\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(C=\dfrac{2x-6\sqrt{x}+7\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

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

\(C=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

\(c.M=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+x}=\dfrac{\sqrt{x}+1+x}{x+\sqrt{x}}.\dfrac{\sqrt{x}+x}{\sqrt{x}}=\dfrac{\sqrt{x}+1+x}{\sqrt{x}}\)

ĐKXĐ: \(x\ne1\), x > 0

\(H=\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+\sqrt{x}}+\dfrac{\sqrt{x^3}-x}{\sqrt{x}-1}\)

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

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

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

\(\left\{{}\begin{matrix}x>1\\H=\dfrac{\sqrt{x-1}+\sqrt{x}}{-1}+\dfrac{\sqrt{x-1}-\sqrt{x}}{-1}+x\end{matrix}\right.\)

\(H=x-2\sqrt{x-1}\)

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

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

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

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

\(=\dfrac{x+3}{\sqrt{x}+3}\)