Question

[( 2 la puterea a 10) la puterea 8 +6 la puterea 5 a •3 la puterea a 7 + 2 la puterea 32] : [(25) la puterea 16+ 6 la puterea a 12: 2 la puterea 7 +29]​

Answer 1

Răspuns:

$\frac{(2^8^0+9*18^5+2^3^2}{25^1^6+32*3^1^2+29}$

Explicație pas cu pas:

[(2¹⁰)⁸+6⁵×3⁷+2³²]:[25¹⁶+6¹²:2⁷+29]=

=(2⁸⁰+3²×18⁵+2³²):(25¹⁶+$\frac{6^1^2}{2^7}$+29)

=2⁸⁰+9×18⁵+2³²):(25¹⁶+$\frac{3^1^2*2^1^2}{2^7}$+29)

=(2⁸⁰+9×18⁵+2³²):(25¹⁶+3¹²×2⁵+29)

=(2⁸⁰+9×18⁵+2³²):(25¹⁶+3¹²×32+29)

=(2⁸⁰+9×18⁵+2³²):(25¹⁶+32×3¹²+29)

=$\frac{(2^8^0+9*18^5+2^3^2}{25^1^6+32*3^1^2+29}$