Question

\(x^2-3x\sqrt{y}+2y\) với x=\(\dfrac{1}{\sqrt{5}-2}\) và y=\(\dfrac{1}{9+4\sqrt{5}}\)

Answer 1

Có : \(x=\dfrac{1}{\sqrt{5}-2}=\dfrac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\)

\(y=\dfrac{1}{9+4\sqrt{5}}=\dfrac{9-4\sqrt{5}}{81-80}=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=\left(\sqrt{5}+2\right)^2=9+4\sqrt{5}\\\sqrt{y}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\end{matrix}\right.\)

Khi đó \(x^2-3x\sqrt{y}+2y=9+4\sqrt{5}-3.\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)+2.\left(9-4\sqrt{5}\right)\)

\(=9+4\sqrt{5}-3\left(5-4\right)+18-8\sqrt{5}\)

\(=24-4\sqrt{5}\)

Answer 2

24 - \(4\sqrt{5}\)