\(R=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge2+1=3\)

ta có \(A=\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

            \(=\sqrt{\frac{1}{x}-\frac{1}{x^2}}+\sqrt{\frac{1}{y}-\frac{2}{y^2}}+\sqrt{\frac{1}{z}-\frac{3}{x^2}}=\sqrt{\frac{1}{4}-\left(\frac{1}{x^2}-2.\frac{1}{2}x+\frac{1}{4}\right)}+\sqrt{\frac{1}{8}-\left(\left(\sqrt{2}y\right)^2-2.\frac{\sqrt{2}}{2\sqrt{2}}x+\frac{1}{8}\right)}+\sqrt{\frac{1}{2}-\left(\left(\sqrt{3}z\right)^2-\frac{1}{z}+\frac{1}{12}\right)}\)

             \(=\sqrt{\frac{1}{4}-\left(\frac{1}{x}-\frac{1}{2}\right)^2}+\sqrt{\frac{1}{8}-\left(\frac{\sqrt{2}}{y}-\frac{1}{2\sqrt{2}}\right)^2}+\sqrt{\frac{1}{12}-\left(\frac{\sqrt{3}}{z}-\frac{1}{2\sqrt{3}}\right)^2}\)

ta có \(\sqrt{\frac{1}{4}-\left(\frac{1}{x}-\frac{1}{2}\right)^2}\le\frac{1}{2}\) ; \(\sqrt{\frac{1}{8}-\left(\frac{\sqrt{2}}{y}-\frac{1}{2\sqrt{2}}\right)^2}\le\frac{1}{2\sqrt{2}}\); \(\sqrt{\frac{1}{12}-\left(\frac{\sqrt{3}}{z}-\frac{1}{2\sqrt{3}}\right)^2}\le\frac{1}{2\sqrt{3}}\)

vậy giá trị lớn nhất của A =\(\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\) khi x=; y=4;z=6

b, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

Ta có : \(B=\dfrac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x+2}{\sqrt{x}}\)

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

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

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

b) Ta có: \(B=\dfrac{x\sqrt{x}-8}{x-2\sqrt{x}}-\dfrac{x\sqrt{x}+8}{x+2\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)

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

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

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

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

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

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

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

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

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

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

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

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

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

Với \(x\ge0;x\ne4\) có:

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

a

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

b

\(P^2=P+2\\ \Leftrightarrow P^2-P-2=0\\ \Leftrightarrow P^2-2P+P-2=0\\ \Leftrightarrow P\left(P-2\right)+\left(P-2\right)=0\\ \Leftrightarrow\left(P-2\right)\left(P+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}P=2\\P=-1\end{matrix}\right.\)

Với P = 2 có:

\(\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}=2\\ \Leftrightarrow2\left(\sqrt{x}+1\right)=4\sqrt{x}+1\\ \Leftrightarrow2\sqrt{x}+2-4\sqrt{x}-1=0\\\Leftrightarrow -2\sqrt{x}+1=0\\\Leftrightarrow-2\sqrt{x}=-1\\\Leftrightarrow \sqrt{x}=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{1}{4} \)

Với P = -1 có:

\(\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}=-1\\ \Leftrightarrow-\sqrt{x}-1-4\sqrt{x}-1=0\\ \Leftrightarrow-5\sqrt{x}=2\\ \Leftrightarrow\sqrt{x}=-\dfrac{2}{5}\left(loại\right)\)

Vậy để \(P^2=P+2\) thì \(x=\dfrac{1}{4}\)

a: P=A:B

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

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

b: P^2=P+2

=>P^2-P-2=0

=>(P-2)(P+1)=0

=>P=2(nhận) hoặc P=-1(loại)

=>\(4\sqrt{x}+1=2\sqrt{x}+2\)

=>2căn x=1

=>x=1/4

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

\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\\ \Rightarrow P=\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{5-\sqrt{5}}{5}\\ c,\dfrac{P}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}\le\dfrac{1}{0-1}=-1\)

Vậy \(\left(\dfrac{P}{\sqrt{x}}\right)_{max}=-1\Leftrightarrow x=0\)

a: ĐKXĐ: x>0; x<>4

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

b: P=2/3

=>(4-căn x)/(căn x-2)=2/3

=>2căn x-4=12-3căn x

=>5căn x=16

=>x=256/25

c: Khi x=8-2căn 7 thì \(P=\dfrac{4-\sqrt{7}+1}{\sqrt{7}-1-2}=\dfrac{5-\sqrt{7}}{\sqrt{7}-3}=-4-\sqrt{7}\)