Câu hỏi của Bùi Đức Anh

Question

Cho x,y >0. Rút gọn A=$\sqrt{2\left(\sqrt{x^2+y^2}+x\right)\left(\sqrt{x^2+y^2}+y\right)}-\sqrt{x^2+y^2}-x-y+2020$

Nguyễn Việt Lâm · Accepted Answer

$A=\sqrt{2\left(x^2+y^2+\left(x+y\right)\sqrt{x^2+y^2}+xy\right)}-\sqrt{x^2+y^2}-x-y+2020$

$=\sqrt{\left(x^2+y^2+2xy\right)+x^2+y^2+2\left(x+y\right)\sqrt{x^2+y^2}}-\sqrt{x^2+y^2}-x-y+2020$

$=\sqrt{\left(x+y\right)^2+2\left(x+y\right)\sqrt{x^2+y^2}+x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020$

$=\sqrt{\left(x+y+\sqrt{x^2+y^2}\right)^2}-\sqrt{x^2+y^2}-x-y+2020$

$=x+y+\sqrt{x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020$

$=2020$Câu trả lời: $A=\sqrt{2\left(x^2+y^2+\left(x+y\right)\sqrt{x^2+y^2}+xy\right)}-\sqrt{x^2+y^2}-x-y+2020$

$=\sqrt{\left(x^2+y^2+2xy\right)+x^2+y^2+2\left(x+y\right)\sqrt{x^2+y^2}}-\sqrt{x^2+y^2}-x-y+2020$

$=\sqrt{\left(x+y\right)^2+2\left(x+y\right)\sqrt{x^2+y^2}+x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020$

$=\sqrt{\left(x+y+\sqrt{x^2+y^2}\right)^2}-\sqrt{x^2+y^2}-x-y+2020$

$=x+y+\sqrt{x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020$

$=2020$

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