\(a,P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}+\dfrac{6\sqrt{x}-4}{1-x}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Tại \(x=7-4\sqrt{3}\) Ta có :
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{7-4\sqrt{3}}-1}{\sqrt{7-4\sqrt{3}}+1}\\ =\dfrac{\sqrt{4-4\sqrt{3}+3}-1}{\sqrt{4-3\sqrt{3}+3}+1}\\ =\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}-1}{\sqrt{\left(2-\sqrt{3}\right)^2}+1}\\ =\dfrac{2-\sqrt{3}-1}{2-\sqrt{3}+1}\\ =\dfrac{1-\sqrt{3}}{3-\sqrt{3}}\\ =\dfrac{1-\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-1\right)}\\ =\dfrac{-1}{\sqrt{3}}\)
