Tính giá trị của biểu thức N= $x^{2019}$ +$3x^{2020}$-$2x^{2021}$ với x= $\frac{√(√5 +2) +√(√5 -2)}{√(√5 +1}$ -√(3 +2√2)
1 câu trả lời
$\displaystyle \begin{array}{{>{\displaystyle}l}} x=\frac{\sqrt{\sqrt{5} +2} +\sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1}} -\sqrt{3+2\sqrt{2}}\\ x=\frac{\left(\sqrt{\sqrt{5} +2} +\sqrt{\sqrt{5} -2}\right)\left(\sqrt{\sqrt{5} +2} -\sqrt{\sqrt{5} -2}\right)}{\left(\sqrt{\sqrt{5} +1}\right)\left(\sqrt{\sqrt{5} +2} -\sqrt{\sqrt{5} -2}\right)} -\sqrt{3+2\sqrt{2}}\\ x=\frac{\sqrt{5} +2 -\sqrt{5} +2}{\sqrt{\sqrt{5} +1} .\sqrt{\sqrt{5} +2} -\sqrt{\sqrt{5} +1} .\sqrt{\sqrt{5} -2}} -\sqrt{1+2\sqrt{2} +2}\\ x=\frac{4}{\sqrt{7+3\sqrt{5}} -\sqrt{3-\sqrt{5}}} -\sqrt{\left( 1+\sqrt{2}\right)^{2}}\\ \frac{1}{\sqrt{2}} x=\frac{4}{\sqrt{14+6\sqrt{5}} -\sqrt{6-2\sqrt{5}}} -\frac{1+\sqrt{2}}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} x=\frac{4}{\sqrt{9+2.3\sqrt{5} +5} -\sqrt{5-2\sqrt{5} +1}} -\frac{1+\sqrt{2}}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} x=\frac{4}{\sqrt{\left( 3+\sqrt{5}\right)^{2}} -\sqrt{\left(\sqrt{5} -1\right)^{2}}} -\frac{1+\sqrt{2}}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} x=\frac{4}{3+\sqrt{5} -\sqrt{5} +1} -\frac{1+\sqrt{2}}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} x=1-\frac{1+\sqrt{2}}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} x=\frac{\sqrt{2} -1-\sqrt{2}}{\sqrt{2}} =\frac{-1}{\sqrt{2}}\\ \rightarrow x=-1\\ N=x^{2019} +3x^{2020} -2x^{2021}\\ N=( -1)^{2019} +3.( -1)^{2020} -2( -1)^{2021}\\ N=-1+3+2=4 \end{array}$