Question-113997

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

Answered by bobhans last updated on 16/Sep/20

c o n s i d e r : \sqrt{2} + \sqrt[4]{2} + 1 = {(\sqrt[4]{2})}^{2} + 1 . (\sqrt[4]{2}) + 1^{2} = \frac{{(\sqrt[4]{2})}^{3} - 1^{3}}{\sqrt[4]{2} - 1}

t h e n \frac{7}{{(\sqrt[4]{2})}^{2} + \sqrt[4]{2} + 1} = \frac{7 (\sqrt[4]{2} - 1)}{(\sqrt[4]{2^{3}}) - 1}

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

\frac{7}{2^{1 / 2} + 2^{1 / 4} + 1} = A + B . 2^{1 / 4} + C . 2^{1 / 2} + D . 2^{3 / 4}

\Leftrightarrow (2^{1 / 2} + 2^{1 / 4} + 1) (A + B . 2^{1 / 4} + C . 2^{1 / 2} + D . 2^{3 / 4}) = 7

\Leftrightarrow A . 2^{1 / 2} + A . 2^{1 / 4} + A + B . 2^{3 / 4} + B . 2^{1 / 2} + B . 2^{1 / 4}

+ 2 C + C . 2^{3 / 4} + C . 2^{1 / 2} + D . 2^{5 / 4} + D . 2 + D . 2^{3 / 3}

= (A + 2 C + 2 D) + (A + B + C) . 2^{1 / 2}

+ (A + B + 2 D) . 2^{1 / 4} + {(B + C + D)}^{3 / 4} = 7

\Leftrightarrow {\begin{cases} A + 2 C + 2 D = 7 (sinceD . 2^{5 / 4} = 2 D . 2^{1 / 4}) \\ A + B + C = 0 \\ A + B + 2 D = 0 \\ B + C + D = 0 \end{cases}

Solve above system we get :

A = 1, B = - 3, C = 2, D = 1

Hence, there are three true options

in available answer