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

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

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

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

\(=\dfrac{-9\sqrt{x}-9}{x-9}\) với \(x\ge0,x\ne9\)

a: \(P=\dfrac{15\sqrt{x}-11+\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11+3x+7\sqrt{x}-6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+21\sqrt{x}-14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

b: Khi x=9 thì \(P=\dfrac{9+21\cdot3-14}{\left(3+3\right)\left(3-1\right)}=\dfrac{29}{6}\)

\(a,A=\dfrac{2\cdot2-4}{2-1}=0\\ b,B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ c,AB=\dfrac{2\sqrt{x}-4}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}-4}{\sqrt{x}+1}=\dfrac{5\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\\ AB=5-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\)

Vì \(\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}>0\) nên \(AB< 5\)

a. \(x=4\Rightarrow A=\dfrac{2.\sqrt{4}-4}{\sqrt{4}-1}=0\)

b. \(\Rightarrow B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(6\sqrt{x}-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

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

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

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

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

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

b) Ta có: \(x=4+2\sqrt{3}\)

\(\Leftrightarrow x=3+2\cdot\sqrt{3}\cdot1+1\)

hay \(x=\left(\sqrt{3}+1\right)^2\)

Thay \(x=\left(\sqrt{3}+1\right)^2\) vào biểu thức \(A=\dfrac{x-1}{\sqrt{x}}\), ta được:

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

\(\Leftrightarrow A=\dfrac{\left(3+2\sqrt{3}\right)\left(\sqrt{3}-1\right)}{2}=\dfrac{3\sqrt{3}-3+6-2\sqrt{3}}{2}\)

\(\Leftrightarrow A=\dfrac{\sqrt{3}+3}{2}\)

Vậy: Khi \(x=4+2\sqrt{3}\) thì \(A=\dfrac{\sqrt{3}+3}{2}\)

Ta có: M=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

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

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

b: \(\sqrt{x}+3>=3\)

=>A<=1

Dấu = xảy ra khi x=0

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

Để P nguyên thì căn x-2\(\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{1;25\right\}\)

a) \(A=\dfrac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

b) \(x=6+4\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{6+4\sqrt{2}}\)

\(=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)

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

\(a,A=\dfrac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\left(x\ge0;x\ne4\right)\\ A=\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\\ b,x=6+4\sqrt{2}=\left(2+\sqrt{2}\right)^2\Leftrightarrow\sqrt{x}=2+\sqrt{2}\\ \Leftrightarrow A=\dfrac{\sqrt{2}+2-4}{\sqrt{2}+2-2}=\dfrac{\sqrt{2}-2}{\sqrt{2}}=1-\sqrt{2}\)