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
