Question

[Nguyễn Minh Huy]

Cho \(\left(x+\sqrt{x^2+2018}\right)\left(y+\sqrt{y^2+2018}\right)=2018\)

Answer 1

[Duyên Phạm<3.03012004]

Ta có: \(\left(x+\sqrt{x^2+2018}\right)\left(y+\sqrt{y^2+2018}\right)=2018\)

\(\Leftrightarrow\left(x+\sqrt{x^2+2018}\right)\left(x-\sqrt{x^2+2018}\right)\left(y+\sqrt{y^2+2018}\right)=2018\left(x-\sqrt{x^2+2018}\right)\)

\(\Leftrightarrow\left(x^2-\left(x+2018\right)^2\right)\left(y+\sqrt{y^2+2018}\right)=2018\left(x-\sqrt{x^2+2018}\right)\)

\(\Leftrightarrow\left(x^2-x^2-2108\right)\left(y+\sqrt{y^2+2018}\right)=2018\left(x-\sqrt{x^2+2018}\right)\)

\(\Leftrightarrow-2018\left(y+\sqrt{y^2+2018}\right)=2018\left(x-\sqrt{x^2+2018}\right)\)

\(\Leftrightarrow-\left(y+\sqrt{y^2+2018}\right)=x-\sqrt{x^2+2018}\)

\(\Leftrightarrow-y-\sqrt{y^2+2018}=x-\sqrt{x^2+2018}\)                 (1)

Và có: \(\left(x+\sqrt{x^2+2018}\right)\left(y+\sqrt{y^2+2018}\right)=2018\)

\(\Leftrightarrow\left(x+\sqrt{x^2+2018}\right)\left(y+\sqrt{y^2+2018}\right)\left(y-\sqrt{y^2+2018}\right)=2018\left(y-\sqrt{y^2+2018}\right)\)

\(\Leftrightarrow\left(x-\sqrt{x^2+2018}\right)\left(y^2-y^2-2018\right)=2018\left(y-\sqrt{y^2+2018}\right)\)

\(\Leftrightarrow-2018\left(x-\sqrt{x^2+2018}\right)=2018\left(y-\left(\sqrt{y^2+2018}\right)\right)\)

\(\Leftrightarrow-x-\sqrt{x^2+2018}=y-\sqrt{y^2+2018}\)                        (2)

Lấy (1) + (2) vế + vế ta được:

\(\left(-y-\sqrt{y^2+2018}\right)+\left(-x-\sqrt{x^2+2018}\right)=\left(x-\sqrt{x^2+2018}\right)+\left(y-\sqrt{y^2+2018}\right)\)

<=>​\(-y-\sqrt{y^2+2018}+-x-\sqrt{x^2+2018}=x-\sqrt{x^2+2018}+y-\sqrt{y^2+2018}\)​

<=> -y - x = x + y

<=> 2y - 2x =0

<=> -2(x+y)=0

<=> x + y =0

vậy x+y=0

cộng điểm cho mk nha!!!!!!!!!!