a, Với x >= 0 ; x khác 4 

\(=\frac{x-3\sqrt{x}+2-\left(x+4\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{-3\sqrt{x}-3-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-7\sqrt{x}-6-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-\sqrt{x}-6}{\sqrt{x}-2}\)

b, \(Q+1>0\Leftrightarrow\frac{-\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}-2}>0\Leftrightarrow\frac{-8}{\sqrt{x}-2}>0\)

\(\Rightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\Rightarrow0\le x< 4\)

c, \(\frac{-\left(\sqrt{x}+6\right)}{\sqrt{x}-2}=\frac{-\left(\sqrt{x}-2+8\right)}{\sqrt{x}-2}=-1-\frac{8}{\sqrt{x}-2}\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

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

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

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 có giá trị là một số nguyên thì:

\(\left(\sqrt{x}+1\right)⋮\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)+4⋮\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow4⋮\left(\sqrt{x}-3\right)\)

Vì \(x\in Z\) nên \(\left(\sqrt{x}-3\right)\inƯ\left(4\right)=\left\{\pm1,\pm2,\pm4\right\}\)

Ta có bảng sau:

Vậy ....

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

Để A có giá trị là một số nguyên khi:

\(4⋮\sqrt{x}-3\) hay \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Do đó:

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

\(\sqrt{x}-3=1\Rightarrow\sqrt{x}=1+3=4\Rightarrow x=16\)

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

\(\sqrt{x}-3=2\Rightarrow\sqrt{x}=2+3=5\Rightarrow x=25\)

\(\sqrt{x}-3=-4\Rightarrow\sqrt{x}=-4+3=-1\)  ( loại )

\(\sqrt{x}-3=4\Rightarrow\sqrt{x}=4+3=7\Rightarrow x=49\)

Vậy để A là một số nguyên khi \(x\in\left\{4;16;1;25;49\right\}\)

\(B=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3\left(\sqrt{x}-1\right)}{x-5\sqrt{x}+6}\left(ĐKXĐ:x\ne4;x\ne9;x\ge0\right)\)

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

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

 \(B< -1\)\(\Leftrightarrow\) \(\frac{1}{3-\sqrt{x}}< -1\)\(\Rightarrow\sqrt{x}-3< 1\Leftrightarrow x< 16\)

Mặt khác : Vì \(B< -1< 0\)nên \(3-\sqrt{x}< 0\Rightarrow x>9\)

Vậy để \(B< -1\)thì \(9< x< 16\)

\(2B\in Z\Leftrightarrow B\in Z\)

\(\Leftrightarrow\frac{1}{3-\sqrt{x}}\in Z\)=> \(3-\sqrt{x}\inƯ\left(1\right)\)

\(\Rightarrow3-\sqrt{x}\in\left\{-1;1\right\}\)\(\Rightarrow x\in\left\{16\right\}\)( Loại x = 4 vì không thoả mãn điều kiện)

Xin lỗi vì để bài mình ghi lộn :))

Còn lại thì ổn rồi :))

Dùng thẳng cô si vào VT luôn cho nhanh :v!

ĐK: \(x,y,z>0\)

Ta có: \(VP=\frac{1}{2}\left(y+3\right)=\frac{y+3}{2}\)

Mặt khác theo cô si,ta có

\(VT\le\frac{1+x}{2}+\frac{1+y-z}{2}+\frac{1+z-x}{2}\)\(=\frac{y+3}{2}=VP\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=1\\y-z=1\\z-x=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y-z=1\\z-1=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\z=2\\y-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=2\end{cases}}\)

Vậy ...

Quá nhanh quá ngu hiểm :v.Lâu lắm mới nghĩ ra được cách thế này.Nãy ngồi bình phương suốt mà làm hoài không ra.

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 \(x\in\left\{0;16\right\}\)