\(a,Q=\dfrac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{x}\left(x>0;x\ne9\right)\\ Q=\dfrac{2x+32}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{x}=\dfrac{2x+32}{x\left(\sqrt{x}+3\right)}\)

a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+3\sqrt{x}+9}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{3x+9}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{3x+9}{x+4\sqrt{x}+3}\)

b: Để A<-1 thì A+1<0

\(\Leftrightarrow\dfrac{3x+9+x+4\sqrt{x}+3}{x+4\sqrt{x}+3}< 0\)

\(\Leftrightarrow\dfrac{4x+4\sqrt{x}+12}{x+4\sqrt{x}+3}< 0\)

hay \(x\in\varnothing\)

\(Q=\dfrac{\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-1-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{-6\sqrt{x}+1+8\sqrt{x}+24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{2\sqrt{x}+25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

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

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

\(a,A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\\ =2.2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\sqrt{3^2}-1}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =-\dfrac{2\left(\sqrt{3}-1\right)}{2}+\left|\sqrt{3}+1\right|\\ =-\sqrt{3}+1+\sqrt{3}+1\\ =2\)

\(B=\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\left(dk:x\ge0,x\ne1\right)\\ =\left(1+\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\right)\left(1-\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\\ =1-x\)

\(b,A=4\sqrt{B}\Leftrightarrow4\sqrt{1-x}=2\\ \Leftrightarrow\sqrt{1-x}=\dfrac{1}{2}\\ \Leftrightarrow\left|1-x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow1-x=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

Vậy \(x=\dfrac{3}{4}\) thì \(A=4\sqrt{B}\).

a) \(A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\)

\(A=2\cdot2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-4\sqrt{5}+\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2}\)

\(A=4\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

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

\(A=-\sqrt{3}+1+\sqrt{3}+1\)

\(A=2\)

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

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

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

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

\(B=1-x\)

b) Ta có: \(A=4\sqrt{B}\)

\(\Rightarrow2=4\sqrt{1-x}\)

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

\(\Leftrightarrow1-x=\dfrac{1}{4}\)

\(\Leftrightarrow x=1-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

a) Ta có:

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

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

\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

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

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

b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)

.....Chưa nghĩ ra....

c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)

Vậy Min P = 0 khi x =9.

k - kb với tớ nhia mn!

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

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)

\(=\dfrac{x}{\sqrt{x}-5}\)

a:

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

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

b: Khi x=7-4căn 3 thì 

\(A=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)

c: A=3

=>căn x-2=1

=>x=9(loại)

\(a,A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(dkxd:x\ne4,x\ge0,x\ne9\right)\)

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

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

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

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

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

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

\(b,x=7-4\sqrt{3}\Rightarrow A=\dfrac{3}{\sqrt{7-4\sqrt{3}}-2}=\dfrac{3}{\sqrt{\left(\sqrt{3}-2\right)^2}-2}=\dfrac{3}{\left|\sqrt{3}-2\right|-2}=\dfrac{3}{-\sqrt{3}+2-2}=\dfrac{\sqrt{3^2}}{-\sqrt{3}}=-\sqrt{3}\)

\(c,A=3\Rightarrow\dfrac{3}{\sqrt{x}-2}=3\\ \Rightarrow\dfrac{3-3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=0\\ \Rightarrow3-3\sqrt{x}+6=0\\ \Rightarrow-3\sqrt{x}=-9\\ \Rightarrow\sqrt{x}=3\\ \Rightarrow x=9\left(ktm\right)\)

Vậy không có giá trị x thỏa mãn đề bài.