This should be easy...

Algebra Level 3

( 2 x − 2 − x − 2 ) 2 + 1 + 1 ( 2 x − 2 − x − 2 ) 2 + 1 − 1 = ( 2 □ + 1 2 □ − 1 ) 2 \large \dfrac{\sqrt{(2^x - 2^{-x - 2})^2 + 1} + 1}{\sqrt{(2^x - 2^{-x - 2})^2 + 1} - 1} = \left(\dfrac{2^\square + 1}{2^\square - 1}\right)^2

Which of the following expression should be put in the box (assuming x x is real)?

x x x − 2 x - 2 x + 2 x + 2 x + 1 x + 1 A different expression. x − 1 x - 1

Thành Đạt Lê by Thành Đạt Lê , 17, Singapore

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1 solution

Chew-Seong Cheong
Jan 29, 2019

X = ( 2 x − 2 − x − 2 ) 2 + 1 + 1 ( 2 x − 2 − x − 2 ) 2 + 1 − 1 Multiply up and down by 2 = ( 2 x + 1 − 2 − ( x + 1 ) ) 2 + 4 + 2 ( 2 x + 1 − 2 − ( x + 1 ) ) 2 + 4 − 2 = 2 2 ( x + 1 ) − 2 + 2 − 2 ( x + 1 ) + 4 + 2 2 2 ( x + 1 ) − 2 + 2 − 2 ( x + 1 ) + 4 − 2 = 2 2 ( x + 1 ) + 2 + 2 − 2 ( x + 1 ) + 2 2 2 ( x + 1 ) + 2 + 2 − 2 ( x + 1 ) − 2 = ( 2 x + 1 + 2 − ( x + 1 ) ) 2 + 2 ( 2 x + 1 + 2 − ( x + 1 ) ) 2 − 2 = 2 x + 1 + 2 − ( x + 1 ) + 2 2 x + 1 + 2 − ( x + 1 ) − 2 Multiply up and down by 2 x + 1 = 2 2 ( x + 1 ) + 2 ⋅ 2 x + 1 + 1 2 2 ( x + 1 ) − 2 ⋅ 2 x + 1 + 1 = ( 2 x + 1 + 1 2 x + 1 − 1 ) 2 \begin{aligned} X & = \frac {\sqrt{(2^x-2^{-x-2})^2+1}+1}{\sqrt{(2^x-2^{-x-2})^2+1}-1} & \small \color{#3D99F6} \text{Multiply up and down by }2 \\ & = \frac {\sqrt{(2^{x+1}-2^{-(x+1)})^2+4}+2}{\sqrt{(2^{x+1}-2^{-(x+1)})^2+4}-2} \\ & = \frac {\sqrt{2^{2(x+1)}-2+2^{-2(x+1)}+4}+2}{\sqrt{2^{2(x+1)}-2+2^{-2(x+1)}+4}-2} \\ & = \frac {\sqrt{2^{2(x+1)}+2+2^{-2(x+1)}}+2}{\sqrt{2^{2(x+1)}+2+2^{-2(x+1)}}-2} \\ & = \frac {\sqrt{(2^{x+1}+2^{-(x+1)})^2}+2}{\sqrt{(2^{x+1}+2^{-(x+1)})^2}-2} \\ & = \frac {2^{x+1}+2^{-(x+1)}+2}{2^{x+1}+2^{-(x+1)}-2} & \small \color{#3D99F6} \text{Multiply up and down by }2^{x+1} \\ & = \frac {2^{2(x+1)} + 2 \cdot 2^{x+1} +1}{2^{2(x+1)} - 2 \cdot 2^{x+1} +1} \\ & = \left(\frac {2^{\boxed{x+1}}+1}{2^{\boxed{x+1}}-1}\right)^2 \end{aligned}

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