a: \(P=1:\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{3x}{2\left(x-4\right)}+\dfrac{2}{2\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{1}{4-2\sqrt{x}}\)
\(=1:\left(\dfrac{2\left(\sqrt{x}-2\right)-3x+2\sqrt{x}+4}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=1:\dfrac{2\sqrt{x}-4-3x+2\sqrt{x}+4}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=\dfrac{2\left(x-4\right)}{-3x+4\sqrt{x}}\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}+2}{3x-4\sqrt{x}}\)
b: Để P=20 thì \(\sqrt{x}+2=60x-80\sqrt{x}\)
\(\Leftrightarrow60x-81\sqrt{x}-2=0\)
Đặt \(\sqrt{x}=a\)
Pt sẽ là \(60a^2-81a-2=0\)
\(\text{Δ}=\left(-81\right)^2-4\cdot60\cdot\left(-2\right)=7041>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}a_1=\dfrac{81-\sqrt{7041}}{120}\left(loại\right)\\a_2=\dfrac{81+\sqrt{7041}}{120}\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\left(\dfrac{81+\sqrt{7041}}{120}\right)^2\)