Question Number 197567 by hardmath last updated on 21/Sep/23
Answered by Frix last updated on 22/Sep/23
$$\mathrm{Obviously}\:{x}={y}=\mathrm{1} \\ $$
Commented by Frix last updated on 22/Sep/23
$$\mathrm{1}^{\mathrm{2}} ×\mathrm{1}+\mathrm{1}×\mathrm{1}^{\mathrm{2}} =\sqrt{\mathrm{1}×\mathrm{1}}+\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{1}+\mathrm{1}=\mathrm{1}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{2}=\mathrm{2} \\ $$$$\sqrt{\mathrm{1}}+\sqrt{\mathrm{1}}+\sqrt{\mathrm{1}×\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{4}=\mathrm{4} \\ $$
Commented by hardmath last updated on 22/Sep/23
$$\mathrm{How} \\ $$