Câu hỏi của Thấu Minh Phong

Question

help me ,pls

Rút gọn rồi tính:

E=$\dfrac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}$, x= 2($\sqrt{3}$+1)

Học tốt · Accepted Answer

ĐK: x$\ge$2

$E=\dfrac{\sqrt{x+2+2\sqrt{\left(x+2\right)\left(x-2\right)}+x-2}}{\sqrt{x^2-4}+x+2}$

$E=\dfrac{\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}}{\sqrt{x^2-4}+x+2}$

$E=\dfrac{\left|\sqrt{x+2}+\sqrt{x-2}\right|}{\sqrt{x^2-4}+x+2}$

$E=\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\left(x+2\right)+\sqrt{\left(x+2\right)\left(\sqrt{x-2}\right)}}$

$E=\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}\left(\sqrt{x+2}+\sqrt{x-2}\right)}$

$E=\dfrac{1}{\sqrt{x+2}}$

Thế x=2($\sqrt{3}+1$)=$2\sqrt{3}+2$ vào E:

=>$E=\dfrac{1}{\sqrt{2\sqrt{3}+4}}$

=>$E=\dfrac{1}{\sqrt{3+2\sqrt{3}+1}}=\dfrac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{1}{\sqrt{3}+1}$

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