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Question Number 113997 by Aina Samuel Temidayo last updated on 16/Sep/20

Answered by bobhans last updated on 16/Sep/20

$$consider:\sqrt{2}+\sqrt[4]{2}+1={\left(\sqrt[4]{2}\right)}^2+1.\left(\sqrt[4]{2}\right)+1^2=\frac{{\left(\sqrt[4]{2}\right)}^3-1^3}{\sqrt[4]{2}-1}$$$$then\frac{7}{{\left(\sqrt[4]{2}\right)}^2+\sqrt[4]{2}+1}=\frac{7(\sqrt[4]{2}-1)}{\left(\sqrt[4]{2^3}\right)-1}$$$$$$

Answered by 1549442205PVT last updated on 16/Sep/20

$$\frac{7}{2^{1/2}+2^{1/4}+1}=\mathrm{A}+\mathrm{B}.2^{1/4}+\mathrm{C}.2^{1/2}+\mathrm{D}.2^{3/4}$$$$\iff (2^{1/2}+2^{1/4}+1)(\mathrm{A}+\mathrm{B}.2^{1/4}+\mathrm{C}.2^{1/2}+\mathrm{D}.2^{3/4})=7$$$$\iff \mathrm{A}.2^{1/2}+\mathrm{A}.2^{1/4}+\mathrm{A}+\mathrm{B}.2^{3/4}+\mathrm{B}.2^{1/2}+\mathrm{B}.2^{1/4}$$$$+2\mathrm{C}+\mathrm{C}.2^{3/4}+\mathrm{C}.2^{1/2}+\mathrm{D}.2^{5/4}+\mathrm{D}.2+\mathrm{D}.2^{3/3}$$$$=(\mathrm{A}+2\mathrm{C}+2\mathrm{D})+(\mathrm{A}+\mathrm{B}+\mathrm{C}).2^{1/2}$$$$+(\mathrm{A}+\mathrm{B}+2\mathrm{D}).2^{1/4}+{(\mathrm{B}+\mathrm{C}+\mathrm{D})}^{3/4}=7$$$$\iff \{\begin{array}{l}\mathrm{A}+2\mathrm{C}+2\mathrm{D}=7(\mathrm{sinceD}.2^{5/4}=2\mathrm{D}.2^{1/4})\\ \mathrm{A}+\mathrm{B}+\mathrm{C}=0\\ \mathrm{A}+\mathrm{B}+2\mathrm{D}=0\\ \mathrm{B}+\mathrm{C}+\mathrm{D}=0\end{array}$$$$\mathrm{Solve}\mathrm{above}\mathrm{system}\mathrm{we}\mathrm{get}:$$$$\mathrm{A}=1,\mathrm{B}=-3,\mathrm{C}=2,\mathrm{D}=1$$$$\mathrm{Hence},\mathrm{there}\mathrm{are}\mathrm{three}\mathrm{true}\mathrm{options}$$$$\mathrm{in}\mathrm{available}\mathrm{answer}$$

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