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Question Number 133469 by Eric002 last updated on 22/Feb/21

$$findxintermsofa$$$$\frac{1+x-\sqrt{2x+x^2}}{1+x+\sqrt{2x+x^2}}=a^3\frac{\sqrt{2+x}+\sqrt{x}}{\sqrt{2+x}-\sqrt{x}}$$

Answered by EDWIN88 last updated on 22/Feb/21

$$\frac{{[(1+\mathrm{x})-\sqrt{2\mathrm{x}+{\mathrm{x}}^2}]}^2}{{(1+\mathrm{x})}^2-(2\mathrm{x}+{\mathrm{x}}^2)}=\frac{{(1+\mathrm{x})}^2+(2\mathrm{x}+{\mathrm{x}}^2)-2(\mathrm{x}+1)\sqrt{2\mathrm{x}+{\mathrm{x}}^2}}{1}$$$$={2\mathrm{x}}^2+4\mathrm{x}+1-2(\mathrm{x}+1)\sqrt{2\mathrm{x}+{\mathrm{x}}^2}$$$$\frac{{[\sqrt{2+\mathrm{x}}+\sqrt{\mathrm{x}}]}^2}{(2+\mathrm{x})-\mathrm{x}}=\frac{2\mathrm{x}+2+2\sqrt{2\mathrm{x}+{\mathrm{x}}^2}}{2}=\mathrm{x}+1+\sqrt{2\mathrm{x}+{\mathrm{x}}^2}$$$$$$$$a^3=\frac{2x^2+4x+1-2(x+1)\sqrt{2x+x^2}}{x+1+\sqrt{2x+x^2}}$$$$a^3=\frac{{[(x+1)-\sqrt{2x+x^2}]}^2}{(x+1)+\sqrt{2x+x^2}}={[(x+1)-\sqrt{2x+x^2}]}^3$$$$\iff \mathrm{x}+1-\sqrt{2\mathrm{x}+{\mathrm{x}}^2}=\mathrm{a}$$$$\Rightarrow {(\mathrm{x}+1)}^2={[\mathrm{a}+\sqrt{2\mathrm{x}+{\mathrm{x}}^2}]}^2$$$$\Rightarrow 1={\mathrm{a}}^2+2\mathrm{a}\sqrt{2\mathrm{x}+{\mathrm{x}}^2}$$$$\Rightarrow {(1-{\mathrm{a}}^2)}^2={4\mathrm{a}}^2(2\mathrm{x}+{\mathrm{x}}^2)$$$$\Rightarrow {\left(\frac{1-{\mathrm{a}}^2}{2\mathrm{a}}\right)}^2+1={(\mathrm{x}+1)}^2$$$$\Rightarrow \mathrm{x}=\sqrt{\frac{{\mathrm{a}}^4+{2\mathrm{a}}^2+1}{{4\mathrm{a}}^2}}-1$$$$\Rightarrow \mathrm{x}=\frac{{\mathrm{a}}^2+1}{2\mathrm{a}}-1=\frac{{\mathrm{a}}^2-2\mathrm{a}+1}{2\mathrm{a}}$$

Commented by Eric002 last updated on 22/Feb/21

$$welldone$$

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