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Question Number 156333 by MathSh last updated on 10/Oct/21

Answered by Rasheed.Sindhi last updated on 10/Oct/21

$$\underset{―}{\mathrm{NOT}\mathrm{General}\mathrm{Solution}}$$$$\mathrm{Simple}\mathrm{special}\mathrm{case}:\mathrm{n}=1$$$${\mathrm{x}}_1=\sqrt{{\mathrm{x}}_1+22}-\sqrt{{\mathrm{x}}_1+1}$$$${\mathrm{x}}_1^2={\mathrm{x}}_1+22+{\mathrm{x}}_1+1-2\sqrt{({\mathrm{x}}_1+22)({\mathrm{x}}_1+1)}\mathrm{y}$$$${2\mathrm{x}}_1+23-{\mathrm{x}}_1^2=2\sqrt{({\mathrm{x}}_1+22)({\mathrm{x}}_1+1)}$$$${({\mathrm{x}}_1^2-{2\mathrm{x}}_1-23)}^2=4({\mathrm{x}}_1^2+{23\mathrm{x}}_1+22)$$$${\mathrm{x}}_1^4+\cancel{{4\mathrm{x}}_1^2}+{23}^2-{4\mathrm{x}}_1^3+\cancel{{92\mathrm{x}}_1}-{46\mathrm{x}}_1^2$$$$-\cancel{{4\mathrm{x}}_1^2}-\cancel{{92\mathrm{x}}_1}-88=0$$$${\mathrm{x}}_1^4-{4\mathrm{x}}_1^3-{46\mathrm{x}}_1^2+441=0$$$$({\mathrm{x}}_1-3)({\mathrm{x}}_1^3-{\mathrm{x}}_1^2-{49\mathrm{x}}_1-147)=0$$$${\mathrm{x}}_1=3\mid {\mathrm{x}}_1^3-{\mathrm{x}}_1^2-{49\mathrm{x}}_1-147=0$$$$Theonlyrationalsolution:{\mathrm{x}}_1=3$$

Commented by MathSh last updated on 10/Oct/21

$$\mathrm{Thank}\mathrm{You}\mathrm{Dear}\mathbf{S}\mathrm{ear}$$$$\mathrm{I}\mathrm{think}\mathrm{that}{\mathrm{x}}_1={\mathrm{x}}_2=...={\mathrm{x}}_{\mathbf{n}}$$

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