[toán 9] giải PT

E

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+$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$

\Leftrightarrow $(x^2-x^2-3)(y+\sqrt{y^2+3})=3(x-\sqrt{x^2+3})$

\Leftrightarrow $y+\sqrt{y^2+3}=\sqrt{x^2+3}-x$

+$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$

\Leftrightarrow $(y^2-y^2-3)(x+\sqrt{x^2+3})=3(y-\sqrt{y^2+3})$

\Leftrightarrow $x+\sqrt{x^2+3}=\sqrt{y^2+3}-y$

Cộng theo vế,đc:

x+y=0
 
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