Question

Va rog,e urgent! dau coroniță !
calculați x la puterea y+y la puterea x pentru: x=8³³:[4³²x2³⁴+(2⁵x2²⁰)⁵:(16x2²³)+(7⁵:7⁵-1)³²x3] y=[(11-0¹¹)x(3³-3²)+1²⁰²⁰]x(3²-2³)-3²x2​

Answer 1

x = 2^99 : (2^64 • 2^34 + 2^125 : 2^27 + 0)
x = 2^99 : (2^98 + 2^98)
x = 2^99 : 2^99
x = 1

y = (11 • 3 + 1) • 1 - 18
y = 34 - 18
y = 16

1^(16+16) = 1^32 = 1

Answer 2

Răspuns:

x=8³³:[4³²·2³⁴+ (2⁵·2²⁰)⁵: (16·2²³)+(7⁵:7⁵- 1)³²·4] =

(2³)³³: [(2²)³²·2³⁴+ (2²⁵)⁵: (2⁴·2²³)+(1-1)³²·4] =

2⁹⁹:(2⁶⁴·2³⁴+2¹²⁵: 2²⁷+0·4) =

2⁹⁹:(2⁹⁸+2⁹⁸) =

2⁹⁹: 2·2⁹⁸=

2⁹⁹: 2⁹⁹=

1

y=[(11-0¹¹)x(3³-3²)+1²°°°]x(3²-2³)-3²x2=

[(11-0)·(27-9)+1]·(9-8)-9x2=

[11·18+1]·1-18=

199-18=

181

x^y+y^x=1¹⁸¹+181¹=1+181=182