\(a,ĐK:x\ge1;x\ne3\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\)

\(a,ĐK:x\ne3;x\ge1\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\\ b,A=4\left(2-\sqrt{3}\right)\\ \Leftrightarrow\sqrt{x-1}+\sqrt{2}=8-4\sqrt{3}\\ \Leftrightarrow\sqrt{x-1}=8-4\sqrt{3}-\sqrt{2}\\ \Leftrightarrow x-1=\left(8-4\sqrt{3}-\sqrt{2}\right)^2\\ \Leftrightarrow x=\left(8-4\sqrt{3}-\sqrt{2}\right)^2+1=...\\ d,A=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow x-1=0\Leftrightarrow x=1\)

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

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

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

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

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

\(A\ge\dfrac{1}{3\sqrt{x}}\Leftrightarrow\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}\ge\dfrac{1}{3\sqrt{x}}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}\ge\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2\le3\)

\(\Rightarrow x\le1\)

Kết hợp ĐKXĐ \(\Rightarrow0< x\le1\)

a: Ta có: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)

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

\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

b: Để \(A\ge0\) thì \(\sqrt{x}-3>0\)

hay x>9

Lời giải:

ĐKXĐ: $x>0; x\neq 4$

Sửa lại đề 1 chút.
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(=\frac{2}{\sqrt{x}+2}\)

\(B=\frac{7}{3}A=\frac{14}{3(\sqrt{x}+2)}\)

Với mọi $x>0$ thì hiển nhiên $B>0$. Mặt khác, $\sqrt{x}+2\geq 2$ nên $B=\frac{14}{3(\sqrt{x}+2)}\leq \frac{14}{6}=\frac{7}{3}$

Vậy $0< B\leq \frac{7}{3}$. $B$ đạt giá trị nguyên thì $B=1;2$

$B=1\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=1$

$\Leftrightarrow x=\frac{64}{9}$ (thỏa mãn)

$B=2\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=2$

$\Leftrightarrow x=\frac{1}{9}$ (thỏa mãn)

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

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

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

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

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

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

2: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Để A là số nguyên thì \(2⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(2\right)\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow x\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;3\right\}\)

Vậy: Để A là số nguyên thì \(x\in\left\{2;3\right\}\)

a: \(B=\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\)

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

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

Khi x=16 thì \(B=\dfrac{2\cdot4+2}{\left(4-2\right)\left(4+2\right)}=\dfrac{10}{2\cdot6}=\dfrac{10}{12}=\dfrac{5}{6}\)

b: P=B/A

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

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

c: P<1

=>P-1<0

=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

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

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

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

=>0<=x<4

mà x nguyên

nên \(x\in\left\{0;1;2;3\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;1;2;3\right\}\)

a, đk: \(x\ge0,x\ne9,x\ne4\)

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

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

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

b,\(Q< -1=>\dfrac{-1}{\sqrt{x}-3}+1< 0< =>\dfrac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(< =>\dfrac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)

\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left[{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\)\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\end{matrix}\right.\)\(< =>9< x< 16\)

c, \(=>2Q=\dfrac{-2}{\sqrt{x}-3}=1+\dfrac{1}{\sqrt{x}-3}\in Z\)

\(< =>\sqrt{x}-3\inƯ\left(1\right)=\left\{\pm1\right\}\)\(=>x\in\left\{16;4\right\}\)(loại 4)

=>x=16

a) \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-3\dfrac{\sqrt{x}-1}{x-5\sqrt{x}+6}\) 

Ta có \(x-5\sqrt{x}+6=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>9\\x>2\end{matrix}\right.\) \(\Leftrightarrow x>9\)

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

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

b) \(Q< -1\Leftrightarrow\dfrac{1}{3-\sqrt{x}}< -1\) \(\Leftrightarrow\dfrac{1}{3-\sqrt{x}}+1< 0\) \(\Leftrightarrow\dfrac{4-\sqrt{x}}{3-\sqrt{x}}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-\sqrt{x}>0\\3-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4-\sqrt{x}< 0\\3-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow9< x< 16\)

Vậy để \(Q< -1\) thì \(S=\left\{x/9< x< 16\right\}\)

c) \(2Q\in Z\Leftrightarrow\dfrac{2}{3-\sqrt{x}}\in Z\)

\(\Rightarrow3-\sqrt{x}\inƯ\left(2\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}3-\sqrt{x}=2\\3-\sqrt{x}=-2\\3-\sqrt{x}=1\\3-\sqrt{x}=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=25\\x=4\\x=16\end{matrix}\right.\)

Kết hợp với ĐKXĐ,ta có để \(2Q\in Z\) thì \(x\in\left\{16;25\right\}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

Ta có: \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{3\sqrt{x}-3}{x-5\sqrt{x}+6}\)

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

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

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

c) Để 2Q là số nguyên thì \(-2⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1\right\}\)