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

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

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

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

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

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

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

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

1:

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

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: 

1:

\(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

3: A nguyên

=>-5căn x-15+17 chia hết cho căn x+3

=>căn x+3 thuộc Ư(17)

=>căn x+3=17

=>x=196

a: Khi x=4 thì \(A=\left(\dfrac{2+2}{2+1}-\dfrac{2\cdot2-2}{2-1}\right)\cdot\left(4-1\right)=\dfrac{1}{3}\cdot3=1\)

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

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

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

a) \(A=\dfrac{3x}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)

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

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

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

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