1x -1 < 3 

<=> x - 1 < 3 

<=> x      < 4  ( nhận )

Vay : x < 4 

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

Để A nguyên thì \(\frac{2}{\sqrt{x}+1}\)phải nguyên suy ra \(\sqrt{x}+1\)là ước của 2

Ta thấy \(Ư\left(2\right)=\left\{1;-1;2;-2\right\}\) mà điều kiện cho \(x\ge0\)và \(x\ne1\)nên \(\sqrt{x}+1\in\left\{1;2\right\}\)

Với \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)(thoải mãn )

Với \(\sqrt{x}+1=2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)(loại)

Vậy x = 0 thì A nguyên

1: ĐKXĐ: x>=0; x<>9

\(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

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

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left.\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)\right.}\)

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

2: Sửa đề: \(x=7+4\sqrt3\)

\(x=7+4\sqrt3\)

\(=4+2\cdot2\cdot\sqrt3+3=\left(2+\sqrt3\right)^2\)

\(P=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\left(2+\sqrt3\right)}{2+\sqrt3-3}=\frac{3\left(2+\sqrt3\right)}{\sqrt3-1}\)

\(=\frac{3\left(2+\sqrt3\right)\left(\sqrt3+1\right)}{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}=\frac{3\left(2\sqrt3+2+3+\sqrt3\right)}{3-1}=\frac{3\left(3\sqrt3+5\right)}{2}=\frac{9\sqrt3+15}{2}\)

3: P<1

=>P-1<0

=>\(\frac{3\sqrt{x}}{\sqrt{x}-3}-1<0\)

=>\(\frac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}<0\)

=>\(\frac{2\sqrt{x}+3}{\sqrt{x}-3}\) <0

=>\(\sqrt{x}-3<0\)

=>\(\sqrt{x}<3\)

=>0<=x<9