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

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

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

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

b) Để \(A< -\dfrac{1}{3}\) thì \(A+\dfrac{1}{3}< 0\)

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

\(\Leftrightarrow\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-6< 0\)

\(\Leftrightarrow x< 36\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 36\\x\ne9\end{matrix}\right.\)

a: 

Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)

\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)

b: P>=1/2

=>P-1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)

=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)

=>\(-\sqrt{x}-15>=0\)

=>\(-\sqrt{x}>=15\)

=>căn x<=-15

=>\(x\in\varnothing\)

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>P>=-2

Dấu = xảy ra khi x=0

\(a,P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ P=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}+3}\\ b,P=\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{0+3}=-1\\ P_{min}=-1\Leftrightarrow x=0\)

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

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

Đề bạn gõ sai, mình có sửa lại r nha

\(a,A=\dfrac{1-\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3}\\ x=5\Leftrightarrow A=\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{3}=\dfrac{5-2\sqrt{5}}{3}\\ c,A=-\dfrac{1}{3}\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=-1\Leftrightarrow x-2\sqrt{x}+1=0\\ \Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow x=1\left(ktm\right)\Leftrightarrow x\in\varnothing\)

\(a,A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\left(dk:x\ge0,x\ne4\right)\\ =\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\\ =\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x-4+10-x}\)

\(=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\sqrt{x}-2}.\dfrac{1}{6}\\ =\dfrac{-6}{\left(\sqrt{x}-2\right).6}\\ =-\dfrac{1}{\sqrt{x}-2}\)
\(b,A>0\Leftrightarrow-\dfrac{1}{\sqrt{x}-2}>0\Leftrightarrow\sqrt{x}-2< 0\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với \(dk:x\ge0,x\ne4\), ta kết luận \(0\le x< 4\)

Mình cần gấp nhớ đừng làm tắt nhé 

A = [√x/(x - 4) + 2/(2 - √x) + 1/(√x + 2)] : [(√x - 2 + (10 - x)/(√x + 2)]

= [√x/(√x - 2)(√x + 2) - 2(√x + 2)/(√x - 2)(√x + 2) + (√x - 2)/(√x - 2)(√x + 2)] : [(x - 4 + 10 - x)/(√x + 2)]

= [√x - 2(√x + 2) + (√x - 2)]/[(√x - 2)(√x + 2)] : 6/(√x + 2)

= (√x - 2√x - 4 + √x - 2)/(√x - 2)(√x + 2)] . (√x + 2)/6

= -1/(√x - 2)

Để A > 0 thì -1/(√x - 2) > 0

√x - 2 < 0

√x < 2

x < 4

Vậy 0 ≤ x < 4 thì A > 0

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

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

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

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

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

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}-\sqrt{x}=3\)

\(\Leftrightarrow2\sqrt{x}=3\)

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

\(\Rightarrow x=\dfrac{9}{4}\)

\(a)E=\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-1\right):\left(\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+2}\right):\dfrac{4-x+x-4-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}+2}:\dfrac{9-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2}{\sqrt{x}+2}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{9-x}\\ =\dfrac{-2\left(\sqrt{x}-3\right)}{9-x}=\dfrac{2\left(\sqrt{x}-3\right)}{x-9}\\ =\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2}{\sqrt{x}+3}\)

\(b)\)E dương

\(\Leftrightarrow E>0\\ \Leftrightarrow\dfrac{2}{\sqrt{x}+3}>0\\ \Leftrightarrow\sqrt{x}+3>0\left(Vì.2>0\right)\\ \Leftrightarrow\sqrt{x}>-3\forall x\in R\\ \Rightarrow x\ge0\)

Kết hợp đk

\(x\ge0;x\ne4;x\ne9\) 

Vậy \(x\ge0;x\ne4;x\ne9\) thì E dương