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Question Number 121446 by mr W last updated on 08/Nov/20

Answered by liberty last updated on 08/Nov/20

$$\mathrm{u}=\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}$$$$\mathrm{u}(\mathrm{y}+\sqrt{1+{\mathrm{y}}^2})=1\Rightarrow \sqrt{1+{\mathrm{y}}^2}=\frac{1}{\mathrm{u}}-\mathrm{y}$$$$\Rightarrow 1+{\mathrm{y}}^2=\frac{1}{{\mathrm{u}}^2}-2\frac{\mathrm{y}}{\mathrm{u}}+{\mathrm{y}}^2$$$$\frac{2\mathrm{y}}{\mathrm{u}}=\frac{1}{{\mathrm{u}}^2}-\mathrm{u}\Rightarrow \mathrm{y}=\frac{1}{2}(\frac{1}{\mathrm{u}}-\mathrm{u})$$$$\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}=-\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}$$$$\mathrm{y}=\frac{1}{2}(-\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}-\mathrm{x}-\sqrt{1+{\mathrm{x}}^2})$$$$\mathrm{y}=-\mathrm{x}\Rightarrow \mathrm{x}+\mathrm{y}=0\wedge {(\mathrm{x}+\mathrm{y})}^2=0$$

Answered by MJS_new last updated on 08/Nov/20

$$0$$$$\mathrm{let}x=\frac{p^2-1}{2p}\wedge y=\frac{q^2-1}{2q}$$$$\Rightarrow pq=1\Rightarrow y=-\frac{p^2-1}{2p}\Rightarrow x+y=0$$

Answered by mr W last updated on 08/Nov/20

$$\sqrt{1+y^2}+y=\sqrt{1+x^2}-x$$$$x+y=\sqrt{1+x^2}-\sqrt{1+y^2}$$$$x^2+y^2+2xy=1+x^2+1+y^2-2\sqrt{(1+x^2)(1+y^2)}$$$$1-xy=\sqrt{(1+x^2)(1+y^2)}$$$$1-2xy+x^2{y}^2=(1+x^2)(1+y^2)$$$$1-2xy+x^2{y}^2=1+x^2+y^2+x^2{y}^2$$$$x^2+y^2+2xy=0$$$$\Rightarrow {(x+y)}^2=0$$

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