Câu hỏi của Hoàng Hải Nguyễn

Question

$\left(3-\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\cdot\sqrt{3-\sqrt{5}}$

Nobi Nobita · Accepted Answer

Vì $9>5$$\Rightarrow\sqrt{9}>\sqrt{5}$$\Rightarrow3>\sqrt{5}$$\Rightarrow3-\sqrt{5}>0$mà $3+\sqrt{5}>0$$\Rightarrow\left(3-\sqrt{5}\right).\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}$$=\sqrt{\left(3-\sqrt{5}\right)^2.\left(3+\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)^2.\left(3-\sqrt{5}\right)}$$=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}$$=\sqrt{\left(9-5\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(9-5\right).\left(3+\sqrt{5}\right)}$$=\sqrt{4.\left(3-\sqrt{5}\right)}+\sqrt{4.\left(3+\sqrt{5}\right)}$$=2.\sqrt{3-\sqrt{5}}+2.\sqrt{3+\sqrt{5}}$Câu trả lời: Vì $9>5$$\Rightarrow\sqrt{9}>\sqrt{5}$$\Rightarrow3>\sqrt{5}$$\Rightarrow3-\sqrt{5}>0$mà $3+\sqrt{5}>0$$\Rightarrow\left(3-\sqrt{5}\right).\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}$$=\sqrt{\left(3-\sqrt{5}\right)^2.\left(3+\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)^2.\left(3-\sqrt{5}\right)}$$=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}$$=\sqrt{\left(9-5\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(9-5\right).\left(3+\sqrt{5}\right)}$$=\sqrt{4.\left(3-\sqrt{5}\right)}+\sqrt{4.\left(3+\sqrt{5}\right)}$$=2.\sqrt{3-\sqrt{5}}+2.\sqrt{3+\sqrt{5}}$

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