a: Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}-1\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-4-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-1\)

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

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

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

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

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

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

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

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

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

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

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

Để A có giá trị nguyên \(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\in Z\) \(\Leftrightarrow2⋮\left(\sqrt{x}-1\right)\)\(\Leftrightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)\(\Leftrightarrow\sqrt{x}\in\left\{2;0;3;-1\right\}\)

Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}\in\left\{2;0;3\right\}\Leftrightarrow x\in\left\{4;0;9\right\}\)

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

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

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

=\(\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}\) 

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

Để A nguyên thì \(\sqrt{x}-1\)∈Ư₍₂₎

⇒x∈{4;0;9}

a) \(ĐK:x\ge0,x\ne1\)

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

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

b) \(P=\dfrac{2\sqrt{x}}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)

Kết hợp với đk:

\(\Rightarrow0\le x< 1\)

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

Đề thiếu rồi bạn

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

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 $x=1$