Question

[8+8×(8^2+8^98÷8^96)]÷129​

Answer 1

$[8+8×(8^2+ {8}^{98} ÷ {8}^{96} )]÷129 =$

$[8+8×(8^2+ {8}^{98 - 96} )]÷129 =$

$[8+8×(8^2+ {8}^{2} )]÷129 =$

$8+8×8^2+ 8 \times {8}^{2} )÷129 =$

$(8+8^3+ {8}^{3} )÷129 =$

$[8×(1 + 8^2+ {8}^{2} )]÷129 =$

$[8×(1 + 64+ 64 )]÷129 =$

$(8×129 )÷129 = \frac{8 \times 129}{129} = 8$

Answer 2

Explicație pas cu pas:

ღSalutare!ღ

[8+8×(8²+8⁹⁸:8⁹⁶)]:129 =

= [8+8×(8²+8^(98-96) )]:129 =

= [8+8×(8²+8²)]:129 =

= [8+8×(64+64)]:129 =

= (8+8×128):129 =

= (8+1024):129 =

= 1032:129 =

= 8

^=la puterea

Succes!:)