\(P=\frac{\sqrt{x}}{\sqrt{x-1}}\)

\(P^2=\frac{x}{x-1}\)

\(P^2=\frac{x-1+1}{x-1}\)

\(P^2=1+\frac{1}{x-1}\)

Để P² nguyên thì \(\frac{1}{x-1}\) nguyên

\(\Rightarrow x-1=1\left(because:x\ge0\right)\)

\(\Rightarrow x=1\)

Thay vào được \(P^2=\frac{1}{0}\left(VL\right)\)

Vậy ko có x thỏa mãn

zZz Cool Kid zZz : đã có x nguyên chưa zậy?

\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để A thuộc Z

=>\(\frac{4}{\sqrt{x}-3}\in Z\)

<=>\(\sqrt{x}-3\inƯ\left(4\right)\)

=>\(\sqrt{x}-3\in\left(-2;2;-1;1;-4;4\right)\)

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

\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\) \(\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)

\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\)\(\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)

\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)

\(E=\frac{x}{\sqrt{x}-1}\)

b) \(E>1\Leftrightarrow\frac{x}{\sqrt{x}-1}>1\)

\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-1>0\)

\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\frac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}-1}>0\)

\(\Rightarrow\sqrt{x}-1>0\)  vì tử của phân số luôn \(\ge0\forall x\ge0\)

\(\Rightarrow x>1\)

kết hợp với ĐKXĐ \(x\ge0\Rightarrow x>1\)

vậy \(x>1\) thì \(E>1\)

a: Để A là số nguyên thì \(\sqrt{x}+1-6⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;2;3;6\right\}\)

hay \(x\in\left\{0;1;2;5\right\}\)

b: 

Để A là số nguyên thì \(\sqrt{x}+1-6⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;2;3;6\right\}\)

hay \(x\in\left\{0;1;2;5\right\}\)

Để \(P\ge1\) thì \(P-1\ge0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}+1}{\sqrt{x}-1}\ge0\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x>1\end{matrix}\right.\)

Kết hợp ĐKXĐ, ta được: x=0 hoặc x>1