\(a,P=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\\ b,P=1\Leftrightarrow\sqrt{x}+1=2\sqrt{x}\\ \Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\\ c,P=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\in Z\\ \Leftrightarrow\sqrt{x}+1⋮2\sqrt{x}\\ \Leftrightarrow2\sqrt{x}+2⋮2\sqrt{x}\\ \Leftrightarrow2\sqrt{x}\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}=1\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=1\)

Biểu thức thiếu dấu. Bạn coi lại.

Lời giải:

a. ĐKXĐ: $x>0$

\(P=\left(\frac{1}{\sqrt{x}(\sqrt{x}+1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}\right):\frac{2}{\sqrt{x}+1}=\frac{1+\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}.\frac{\sqrt{x}+1}{2}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)

b. \(P=1\Leftrightarrow \frac{\sqrt{x}+1}{2\sqrt{x}}=1\Leftrightarrow \sqrt{x}+1=2\sqrt{x}\Leftrightarrow \sqrt{x}=1\Leftrightarrow x=1\) (tm)

c.

\(\frac{\sqrt{x}+1}{2\sqrt{x}}\in\mathbb{Z}\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}}\in\mathbb{Z}\)

\(\Leftrightarrow 1+\frac{1}{\sqrt{x}}\in\mathbb{Z}\Leftrightarrow \frac{1}{\sqrt{x}}\in\mathbb{Z}\)

Với $x$ nguyên thì \(\Rightarrow \sqrt{x}\) là ước của $1$

$\Rightarrow \sqrt{x}\in \left\{1\right\}$

$\Rightarrow x\in\left\{1\right\}$

Thử lại thấy thỏa mãn. Vậy $x=1$ 

Lời giải:

a. ĐKXĐ: $x\geq 0$

$P< \frac{1}{2}\Leftrightarrow \frac{\sqrt{x}}{\sqrt{x}+2}< \frac{1}{2}$

$\Leftrightarrow \frac{\sqrt{x}}{\sqrt{x}+2}-\frac{1}{2}<0$

$\Leftrightarrow \frac{\sqrt{x}-2}{2(\sqrt{x}+2)}<0$

$\Leftrightarrow \sqrt{x}-2<0$ (do mẫu dương rồi) 

$\Leftrightarrow 0\leq x< 4$

Kết hợp đkxđ suy ra $0\leq x< 4$

b. 

Với $x\geq 0$ thì $P\geq 0$

Lại có: $P<1$ (do tử nhỏ hơn mẫu)

$\Rightarrow P$ nguyên khi mà $P=0$

$\Leftrightarrow x=0$

a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)

a: Thay x=36 vào B, ta được:

\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

\(B=\dfrac{x-3}{x-1}-\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)

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

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

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

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

\(ĐặtP=\dfrac{A}{B}\)

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

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

Để P<1 thì P-1<0

=>\(\dfrac{2\sqrt{x}-2-\sqrt{x}}{\sqrt{x}}< 0\)

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

=>\(\sqrt{x}< 2\)

=>0<=x<4

mà x nguyên

nên \(x\in\left\{0;1;2;3\right\}\)

\(P=\dfrac{A}{B}=\sqrt{x}+1\)

P<7/4

=>căn x<3/4

=>0<x<9/16

\(B=\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

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

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

Để B nguyên thì \(\sqrt{x}-3\in\left\{1;-1;5\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;8\right\}\)

hay \(x\in\left\{16;4;64\right\}\)

\(P=\dfrac{B}{A}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}:\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\\ =\dfrac{\sqrt{x}-1-2}{\sqrt{x}-1}\\ =1-\dfrac{2}{\sqrt{x}-1}\)

Để \(P=\dfrac{B}{A}\)  có giá trị nguyên

Thì \(2⋮\left(\sqrt{x}-1\right)\Rightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)=\left\{2;-2;1;-1\right\}\)

Vậy \(x\in\left\{9;4;0\right\}\) thì \(x\) nguyên và \(P\) có giá trị nguyên

đk x > 0 

\(\dfrac{A}{B}=\dfrac{\dfrac{x+2\sqrt{x}}{x}}{\dfrac{\sqrt{x}+2}{\sqrt{x}+1}}=\dfrac{\dfrac{\sqrt{x}+2}{\sqrt{x}}}{\dfrac{\sqrt{x}+2}{\sqrt{x}+1}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{7}{4}< 0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}+4-7\sqrt{x}}{4\sqrt{x}}< 0\Leftrightarrow\dfrac{-3\sqrt{x}+4}{4\sqrt{x}}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3\sqrt{x}+4\ne0\\-3\sqrt{x}+4< 0\\4\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{16}{9}\\x< \dfrac{16}{9}\\x\ne0\end{matrix}\right.\)