\(a,A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1;x\ne9\right)\\ A=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(b,A\in Z\Leftrightarrow\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}\in Z\Leftrightarrow1+\dfrac{5}{\sqrt{x}-3}\in Z\\ \Leftrightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ Mà.x\ge0\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;8\right\}\\ \Leftrightarrow x\in\left\{4;16;64\right\}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne1\end{matrix}\right.\)

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

b) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\in Z\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Kết hợp đk

\(\Rightarrow x\in\left\{4;16;64\right\}\)

Để A nguyên thì \(2\sqrt{x}+3⋮3\sqrt{x}-1\)

\(\Leftrightarrow6\sqrt{x}+9⋮3\sqrt{x}-1\)

\(\Leftrightarrow3\sqrt{x}-1\in\left\{-1;1;11\right\}\)

\(\Leftrightarrow3\sqrt{x}\in\left\{0;12\right\}\)

hay \(x\in\left\{0;16\right\}\)

\(Q=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

b.\(Q< 1\)

\(\Leftrightarrow x-\sqrt{x}-2< x-5\sqrt{x}+6\)

\(\Leftrightarrow4\sqrt{x}-8< 0\)

\(\Leftrightarrow0\le x< 4\)

Vay de Q<1 thi \(0\le0< 4\)

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+8}\left(x\ge0\right)=\dfrac{\sqrt{x}+8-10}{\sqrt{x}+8}=1-\dfrac{10}{\sqrt{x}+8}\)

Để biểu thức nguyên thì \(10⋮\sqrt{x}+8\)

\(\Leftrightarrow\sqrt{x}+8\inƯ\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-18;-13;-10;-9;-7;-6;-3;2\right\}\)

Mà \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}\in\left\{2\right\}\Leftrightarrow x=4\)

Với \(x\ge0\) có : \(\dfrac{\sqrt{x}-2}{\sqrt{x}+8}=\dfrac{\sqrt{x}+8-10}{\sqrt{x}+8}=1-\dfrac{10}{\sqrt{x}+8}\)

Để bthuc nhận gt nguyên thì :

\(\dfrac{10}{\sqrt{x}+8}\in Z\Leftrightarrow10⋮\sqrt{x}+8\) \(\Leftrightarrow\sqrt{x}+8\inƯ\left(10\right)\Leftrightarrow\sqrt{x}+8=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

Mà \(\sqrt{x}\ge0\) nên ta chỉ xét \(\sqrt{x}+8=\left\{1;2;5;10\right\}\)

Ta có bảng sau

Vậy x=4 là gtri cần tìm

Để biểu thức nguyên thì \(3⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2=3\)

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

hay x=1

\(\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

mà \(x>0=>\sqrt{x}+2>2\) nên \(\sqrt{x}+2=\left\{3\right\}=>x=1\left(tm\right)\)

Vaayy.....

Để biểu thức \(\dfrac{3}{\sqrt{x}+2}\) nguyên thì \(3⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2=3\)

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

hay x=1

a: ĐKXĐ: x>=0; x<>25

Sửa đề: \(Q=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)

b: Q=-3/7

=>\(\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=-\dfrac{3}{7}\)

=>7căn x-35=-3căn x-15

=>10căn x=20

=>x=4

c: Q nguyên

=>căn x+5-10 chia hết cho căn x+5

=>căn x+5 thuộc {5;10}

=>căn x thuộc {0;5}

Kết hợp ĐKXĐ, ta được: x=0

a) \(Q=\dfrac{\sqrt[]{x}}{\sqrt[]{x}-5}-\dfrac{10\sqrt[]{x}}{x-25}-\dfrac{5}{\sqrt[]{x}-5}\left(1\right)\)

Q có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-25\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow Q=\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}-5}-\dfrac{10\sqrt[]{x}}{x-25}\)

\(\Leftrightarrow Q=1-\dfrac{10\sqrt[]{x}}{x-25}\)

\(\Leftrightarrow Q=\dfrac{x+10\sqrt[]{x}-25}{x-25}\)

\(\Leftrightarrow Q=\dfrac{\left(\sqrt[]{x}-5\right)^2}{\left(\sqrt[]{x}-5\right)\left(\sqrt[]{x}+5\right)}=\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}+5}\)

b) \(Q=-\dfrac{3}{7}\)

\(\Leftrightarrow\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}+5}=-\dfrac{3}{7}\)

\(\Leftrightarrow7\left(\sqrt[]{x}-5\right)=-3\left(\sqrt[]{x}+5\right)\)

\(\Leftrightarrow7\sqrt[]{x}-35=-3\sqrt[]{x}-15\)

\(\Leftrightarrow10\sqrt[]{x}=20\)

\(\Leftrightarrow\sqrt[]{x}=2\Leftrightarrow x=4\)

c) \(Q\in Z\Leftrightarrow\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}+5}\in Z\) \(\left(x\in Z^+\right)\)

\(\Leftrightarrow\sqrt[]{x}-5⋮\sqrt[]{x}+5\)

\(\Leftrightarrow\sqrt[]{x}-5-\left(\sqrt[]{x}-5\right)⋮\sqrt[]{x}-5\)

\(\Leftrightarrow\sqrt[]{x}-5-\sqrt[]{x}-5⋮\sqrt[]{x}+5\)

\(\Leftrightarrow-10⋮\sqrt[]{x}+5\)

\(\Leftrightarrow\sqrt[]{x}+5\in U\left(10\right)=\left\{1;2;5;10\right\}\)

\(\Leftrightarrow x\in\left\{0;25\right\}\)

\(a,ĐK:x>0;x\ne1\\ b,B=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ c,B=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\in Z\\ \Leftrightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{2;3\right\}\left(x>0\right)\Leftrightarrow x\in\left\{4;9\right\}\left(tm\right)\)