1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

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

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

Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)

\(\Leftrightarrow x+2\sqrt{x}-3=0\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow x=1\left(nhận\right)\)

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

\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)

9.

\(A>1\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-1}>1\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-1>0\)

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

\(\Leftrightarrow\dfrac{-1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp với điều kiện giả thiết.

10.

\(P< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 1\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

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

\(\Leftrightarrow\sqrt{x}-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp với điều kiện giả thiết.

Bài 9:

Để A>1 thì A-1>0

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

\(\Leftrightarrow\dfrac{-1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

Bài 10:

Để P<1 thì P-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

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

\(\Leftrightarrow\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

6: Để P>1 thì P-1>0

\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}>0\)

\(\Leftrightarrow\sqrt{a}-2< 0\)

hay a<4

Kết hợp ĐKXĐ, ta được: \(0\le a< 4\)

5: Để P>0 thì \(x-4\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}-4>0\)

hay x>16

\(A=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}+\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\left(x>0,x\ne1\right)\)

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

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

\(A-\left|A\right|=0\Rightarrow A=\left|A\right|\Rightarrow\left[{}\begin{matrix}A=A\\A=-A\end{matrix}\right.\)

mà \(x>0\Rightarrow A>0\Rightarrow A=A\) (luôn đúng với mọi \(x\in R\) )

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

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

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

Để A-|A|=0 thì A=|A|

\(\Leftrightarrow3x+3\ge0\)

hay \(x\ge-1\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}=\dfrac{x-1}{\sqrt{x}}+\dfrac{5}{\sqrt{x}}=\dfrac{x+4}{\sqrt{x}}\)

b: Để A=5 thì \(x+4=5\sqrt{x}\)

=>x=1(loại) hoặc x=16(nhận)

1: Để A>0 thì x-1<0

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

1) Để A > 0 thì:

\(x-1< 0\Leftrightarrow x< 1\)

\(\Rightarrow0\le x< 1\) và \(x\ne1\)

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

Để A<1 thì \(\dfrac{2}{\sqrt{x}-1}< 0\)

\(\Rightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)

Mà x\(\ge0,x\ne1\)

\(\Rightarrow0\le x< 1\)

Bài 2: 

Để A<1 thì A-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

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

\(\Leftrightarrow\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

Em kiểm tra lại đề, mẫu số của phân số đầu tiên chắc chắn bị sai

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

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

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

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

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

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

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

\(---\)

\(b,A< 0\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

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

\(\Leftrightarrow x< 1\)

Kết hợp với điều kiện của \(x\), ta được:

\(0\le x< 1\)

Vậy: ...

\(Toru\)

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

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

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

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

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

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

b) \(A< 0\) khi

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

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

\(\Leftrightarrow x< 1\)

Kết hợp với đk:

\(0\le x< 1\)

(a) Với \(x\ge0,x\ne9\), ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

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

\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{3}{\sqrt{x}+3}.\)

(b) Ta có: \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)

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

Thay vào biểu thức \(A\) (thỏa mãn điều kiện), ta được: \(A=\dfrac{3}{2+\sqrt{3}+3}=\dfrac{3}{5+\sqrt{3}}\)

\(=\dfrac{3\left(5-\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{15-3\sqrt{3}}{22}.\)

(c) Để \(A=\dfrac{3}{5}\Rightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{3}{5}\)

\(\Rightarrow\sqrt{x}+2=5\Leftrightarrow x=9\) (không thỏa mãn).

Vậy: \(x\in\varnothing.\)

(d) Để \(A>1\Leftrightarrow A-1>0\Rightarrow\dfrac{3}{\sqrt{x}+3}-1>0\)

\(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}+3}>0\Rightarrow1-\sqrt{x}>0\) (do \(\sqrt{x}+3>0\forall x\inĐKXĐ\))

\(\Rightarrow x< 1\). Kết hợp với điều kiện thì \(0\le x< 1.\)

(e) \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(VL\right)\\\sqrt{x}=-4\left(VL\right)\\\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\\\sqrt{x}=-6\left(VL\right)\end{matrix}\right.\)

Vậy: \(x=0.\)