a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

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

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

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

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

a) ĐKXĐ : \(x\sqrt{x}-1\ge0\Leftrightarrow x\ge1\)

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

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

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

c) Có : \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\)

Khi đó B = \(\dfrac{\sqrt{3}-1}{2}-1=\dfrac{\sqrt{3}-3}{2}\)

\(a,\) B có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

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

\(c,x=\dfrac{2-\sqrt{3}}{2}\Rightarrow B=\sqrt{\dfrac{2-\sqrt{3}}{2}}-1\)

\(=\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}.\sqrt{2}}-\sqrt{2}\) (Nhân \(\sqrt{2}\) để khử căn dưới mẫu)

\(=\dfrac{\sqrt{4-2\sqrt{3}}-2\sqrt{2}}{2}\)

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

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

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

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

a: Thay x=36 vào B, ta được:

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

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

\(a,\) Rút gọn 

\(A=\dfrac{3}{\sqrt{7}-2}+\sqrt{\left(\sqrt{7}-3\right)^2}\)

\(=\dfrac{3}{\sqrt{7}-2}+\left|\sqrt{7}-3\right|\)

\(=\dfrac{3}{\sqrt{7}-2}+3-\sqrt{7}\)

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

\(=\dfrac{3+3\sqrt{7}-6-7+2\sqrt{7}}{\sqrt{7}-2}\)

\(=\dfrac{5\sqrt{7}-10}{\sqrt{7}-2}\)

\(=\dfrac{5\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)

\(=5\)

Vậy \(A=5\)

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1}\left(dkxd:x\ge0,x\ne1\right)\)

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

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

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

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

Vậy \(B=\sqrt{x}-1\)

\(b,\) Để \(B< A\) thì \(\sqrt{x}-1< 5\)

\(\Leftrightarrow\sqrt{x}< 6\)

\(\Leftrightarrow x< 36\)

a) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{x-1}\right):\left(\dfrac{2x}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

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

b) \(A=2\Rightarrow\dfrac{-1}{\sqrt{x}-1}=2\Rightarrow-1=2\sqrt{x}-2\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{4}\)

Lời giải:

ĐK: $x\geq 0; x\neq 1$

a. 

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

\(=\frac{-\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{x-\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{-\sqrt{x}}{x-\sqrt{x}}=\frac{-\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}=\frac{1}{1-\sqrt{x}}\)

b.

$A=2\Leftrightarrow 1-\sqrt{x}=\frac{1}{2}$

$\Leftrightarrow \sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}$ (tm)

A=(\(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{x-1}\) )\(\div\left(\dfrac{2x}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) (ĐK  \(x\ge0,x\ne1\) )

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

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

  =\(\dfrac{-1}{\sqrt{x}-1}\)

câu b

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

a) \(P=\dfrac{x^2-\sqrt[]{x}}{x+\sqrt[]{x}+1}-\dfrac{2x+\sqrt[]{x}}{\sqrt[]{x}}+\dfrac{2\left(x+\sqrt[]{x}-2\right)}{\sqrt[]{x}-1}\)

Điều kiện xác định \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{\sqrt[]{x}\left[\left(\sqrt[]{x}\right)^3-1\right]}{x+\sqrt[]{x}+1}-\dfrac{\sqrt[]{x}\left(2\sqrt[]{x}+1\right)}{\sqrt[]{x}}+\dfrac{2\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+2\right)}{\sqrt[]{x}-1}\)

\(\Rightarrow P=\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-1\right)\left(x+\sqrt[]{x}+1\right)}{x+\sqrt[]{x}+1}-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)

\(\Rightarrow P=\sqrt[]{x}\left(\sqrt[]{x}-1\right)-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)

\(\Rightarrow P=x-\sqrt[]{x}-2\sqrt[]{x}-1+2\sqrt[]{x}+4\)

\(\Rightarrow P=x-\sqrt[]{x}+3\)

b) \(A=\dfrac{P}{2012\sqrt[]{x}}=\dfrac{x-\sqrt[]{x}+3}{2012\sqrt[]{x}}\)\(\)

\(=\dfrac{x-\sqrt[]{x}+\dfrac{1}{4}-\dfrac{1}{4}+3}{2012\sqrt[]{x}}\)

\(=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}{2012\sqrt[]{x}}\)

\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{\dfrac{11}{4}}{2012\sqrt[]{x}}=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\)

Ta lại có  \(\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}\ge0,\forall x\ne0\)

\(\dfrac{1}{\sqrt[]{x}}>0\Rightarrow\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{4.2012}=\dfrac{11}{8048}\)

\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{8048}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt[]{x}=1\Leftrightarrow x=1\)

Vậy \(GTNN\left(A\right)=\dfrac{11}{8048}\left(tạix=1\right)\)

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

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

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

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

b) \(\dfrac{P}{2012\sqrt{x}}=\dfrac{x-\sqrt{x}+3}{2012\sqrt{x}}=\dfrac{\sqrt{x}}{2012}-\dfrac{1}{2012}+\dfrac{3}{2012\sqrt{x}}\)

\(=\left(\dfrac{\sqrt{x}}{2012}+\dfrac{3}{2012\sqrt{x}}\right)-\dfrac{1}{2012}\)

\(\ge2\sqrt{\dfrac{\sqrt{x}.3}{2012^2\sqrt{x}}}-\dfrac{1}{2012}\) (BĐT Cauchy)

\(=\dfrac{2\sqrt{3}}{2012}-\dfrac{1}{2012}=\dfrac{2\sqrt{3}-1}{2012}\)

Dấu "=" xảy ra khi \(\dfrac{\sqrt{x}}{2012}=\dfrac{3}{2012\sqrt{x}}\Leftrightarrow x=3\)(tm)

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 \(x\in\left\{4;0;9\right\}\)