\(A=\dfrac{5-2\sqrt{5}}{\sqrt{5}}-\left(2\sqrt{5}-3\right)+\sqrt{80}\\ A=\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}}-2\sqrt{5}+3+4\sqrt{5}\\ A=\sqrt{5}-2-2\sqrt{5}+3+4\sqrt{5}\\ A=3\sqrt{5}+1\)
\(B=\dfrac{x+\sqrt{x}}{\sqrt{x}}+\dfrac{x-4}{\sqrt{x}+2}\left(ĐKXĐ:x>0\right)\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x+2}}\\ B=\sqrt{x}+1+\sqrt{x}-2\\ B=2\sqrt{x}-1\)
Khi x > 0, A = B
\(\Leftrightarrow2\sqrt{x}-1=3\sqrt{5}+1\\ \Leftrightarrow2\sqrt{x}=3\sqrt{5}+2\\ \Leftrightarrow\sqrt{x}=\dfrac{3\sqrt{5}+2}{2}\\ \Leftrightarrow x=\left(\dfrac{3\sqrt{5}+2}{2}\right)^2=\dfrac{49+12\sqrt{5}}{4}\)
Kết hợp ĐK. Vậy x > 0 thì \(A=B\Leftrightarrow x=\dfrac{49+12\sqrt{5}}{4}\)
