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\(A=x^2-3x\sqrt{y}+2y\) với \(x=\frac{1}{\sqrt{5}-2}\) và \(y=\frac{1}{\sqrt{9+4\sqrt{5}}}\)
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5 tháng 2 2022
Có :
\(x=\dfrac{1}{\sqrt{5}-2}\Rightarrow x^2=\dfrac{1}{\left(\sqrt{5}-2\right)^2}=\dfrac{1}{5-4\sqrt{5}+4}\\ =\dfrac{1}{9-4\sqrt{5}}\\ y=\dfrac{1}{5+4\sqrt{5}}=\dfrac{1}{5+4\sqrt{5}+2}=\dfrac{1}{\left(\sqrt{5}+2\right)^2}\\ \Rightarrow\sqrt{y}=\sqrt{\dfrac{1}{\left(\sqrt{5}+2\right)^2}}=\dfrac{1}{\sqrt{5}+2}\)
\(\Rightarrow A=\dfrac{1}{9-4\sqrt{5}}-3.\dfrac{1}{\sqrt{5}-2}.\dfrac{1}{\sqrt{5}+2}+\dfrac{2}{9+4\sqrt{5}}\\ =\dfrac{1}{9-4\sqrt{5}}-\dfrac{3}{5-4}+\dfrac{2}{9+4\sqrt{5}}\\ =\dfrac{9+\sqrt{5}+2\left(9-4\sqrt{5}\right)}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-3=\dfrac{27-4\sqrt{5}}{81-80-3}\\ =27-4\sqrt{5}-3=24-4\sqrt{5}\)
18 tháng 6 2019
a, \(A=x^2-x\sqrt{y}-2x\sqrt{y}+2y\)
\(=x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-2\sqrt{y}\right)\left(x-\sqrt{y}\right)\)
18 tháng 6 2019
\(a,\)\(A=x^2-3x\sqrt{y}+2y\)
\(=x^2-2x\sqrt{y}-x\sqrt{y}+2y\)
\(=x\left(x-2\sqrt{y}\right)-\sqrt{y}\left(x-2\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(x-2\sqrt{y}\right)\)
\(b,\)Ta có : \(x=\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\)
\(y=\frac{1}{9+4\sqrt{5}}=\frac{9-4\sqrt{5}}{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}=\frac{9-4\sqrt{5}}{81-80}=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(\Rightarrow A=\left[\sqrt{5}+2-\sqrt{\left(\sqrt{5}-2\right)^2}\right]\left[\sqrt{5}+2-2\sqrt{\left(\sqrt{5}-2\right)^2}\right]\)
\(=\left(\sqrt{5}+2-\sqrt{5}-2\right)\left(\sqrt{5}+2-2\sqrt{5}+4\right)\)
\(=4\left(6-\sqrt{5}\right)\)
\(=24-4\sqrt{5}\)
\(y=\frac{1}{\sqrt{9+4\sqrt{5}}}=\frac{1}{\sqrt{\left(\sqrt{5}+2\right)^2}}=\frac{1}{\sqrt{5}+2}\)
\(A=x^2-3x\sqrt{y}+2y=\left(x-2\sqrt{y}\right)\left(x-\sqrt{y}\right)\)
\(=\left(\frac{1}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}\right)\left(\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\right)\)
\(=\left[\frac{\sqrt{5}+2-2\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\right]\left[\frac{\sqrt{5}+2-\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\right]\)
\(=\left[\frac{-\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\right]\left(\frac{4}{5-4}\right)\)
\(=\left(\frac{-1}{\sqrt{5}-2}\right).4=\frac{-4}{\sqrt{5}-2}\)