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ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

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

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

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

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

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

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

=> \(\frac{-3}{\sqrt{x}+3}\div\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{-3}{\sqrt{x}+3}\times\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{4-x}\)

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

a, \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)

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

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

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

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

Khi đó \(P=-x+\sqrt{x}=-\left(7-4\sqrt{3}\right)+\left(2-\sqrt{3}\right)=-5+3\sqrt{3}\)

c, \(P=-x+\sqrt{x}=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

\(\Rightarrow MaxP=\frac{1}{4}\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)

\(\Leftrightarrow\sqrt{x}+1\ge1\)

\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)

\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)

\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)

\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi x=0

Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0

Lời giải:

a) \(A=4\sqrt{x}-\frac{(\sqrt{x}+3)^2(\sqrt{x}-3)}{x-9}=4\sqrt{x}-\frac{(\sqrt{x}+3)(x-9)}{x-9}=4\sqrt{x}-(\sqrt{x}+3)\)

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

b)

\(B=\frac{\sqrt{9x^2+12x+4}}{3x+2}=\frac{\sqrt{(3x)^2+2.3x.2+2^2}}{3x+2}=\frac{\sqrt{(3x+2)^2}}{3x+2}=\frac{|3x+2|}{3x+2}\)

\(B=1\) nếu $x>\frac{-2}{3}$

$B=-1$ nếu $x< \frac{-2}{3}$

\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)

\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)

\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)

\(B=3-\sqrt{x}-\sqrt{x}+3-6\)

\(B=-2\sqrt{x}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3}{\sqrt{x}-6}\)