Question Number 161061 by cortano last updated on 11/Dec/21
Answered by MJS_new last updated on 11/Dec/21
$$\sqrt{{A}}−\sqrt{{B}}={x}\sqrt{{C}} \\ $$$${A}+{B}−\mathrm{2}\sqrt{{AB}}={x}^{\mathrm{2}} {C} \\ $$$$\mathrm{2}\left(\mathrm{2}^{\mathrm{3}/\mathrm{4}} −\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\right)={x}^{\mathrm{2}} \left(\mathrm{2}^{\mathrm{3}/\mathrm{4}} −\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\right) \\ $$$${x}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}=\sqrt{\mathrm{2}}\:\mathrm{because}\:\mathrm{obviously}\:\sqrt{{A}}−\sqrt{{B}}>\mathrm{0} \\ $$