Question

Tinhs giá trị biểu thức

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

Answer 1

\(y=\frac{1}{\sqrt{9+4\sqrt{5}}}=\frac{1}{\sqrt{\left(\sqrt{5}+2\right)^2}}=\frac{1}{\sqrt{5}+2}\)

\(A=x^2-3x\sqrt{y}+2y=\left(x-2\sqrt{y}\right)\left(x-\sqrt{y}\right)\)

\(=\left(\frac{1}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}\right)\left(\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\right)\)

\(=\left[\frac{\sqrt{5}+2-2\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\right]\left[\frac{\sqrt{5}+2-\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\right]\)

\(=\left[\frac{-\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\right]\left(\frac{4}{5-4}\right)\)

\(=\left(\frac{-1}{\sqrt{5}-2}\right).4=\frac{-4}{\sqrt{5}-2}\)