jaqueline j. lugo matemática de nivelación 2 b

1) 15- (−58 )−2

+ 64

−9+( 2−6−96−(−74 ))

−1

+(3√ 34 + 12 )

3

−(−45 +2)=¿

1

(−58 )2 +1296−9

∗( 2∗12∗−3−96−(−74 ))

−1

+(3√ 3+24 )3

−(−45 + 2∗55 )=¿

11∗52

82

−144 (−13 −32+ 74 )

−1

+( 3√ 3+24 )3

−−4+105

=¿

6425−144 (−(1∗2 )−(3∗3 )

6 +74 )

−1

+( 3√53√4 )3

−65=¿

6425

−144 (−116 + 74 )

−1

+(3√53√4

∗3√42

3√42 )3

−65=¿

6425

−144 (−22+2112 )−1

+( 3√5∗164 )3

−65=¿

6425

−144 ( 1−112 )+( 3√804 )3

−65=¿

6425

−144+1 (−1∗12 )+( 3√22∗102∗2 )3

−65=¿

6425

−144−12+( 3√10 )3

23−65=¿

64−144∗2525

−12+108

−65=¿

−353625

−12+ 54−65=¿

−3536−12∗2525

+ 54−65=¿

−383625

+ 54−65=¿

(−3836∗4 )+ (5∗25 )100

−65=

−15219−(6∗20 )100

=¿

−15339100 =153,44

2) 15-

(0,036−1∗102∗10−3 )2∗0,1296∗0,0063

16∗36

−1∗104∗0,6∗10=¿

(( 361000 )−1

∗102∗10−3)2

∗( 129610000 )∗( 61000 )

3

(16 )∗136

∗104∗6

=¿

(( 361000 )−1

∗102∗10−3)2

1∗1296

10000∗( 61000 )

3

16∗1

36∗104∗6

=¿

(( 361000 )−1

∗102∗10−3)2

∗1296

10000∗( 61000 )

3

1104∗663

=¿

(( 361000 )−1

∗102∗10−3)2

∗1296∗( 61000 )

3

10000∗63

104∗6=¿

16∗((( 62103 )−1

∗102∗10−3)2

∗81( 6103 )3)

16∗625∗63

104∗6=¿

( 103

62∗102

103 )2

∗81( 63

109 )625

∗63

104∗6=¿

104

64∗81∗63

109

625∗63

104∗6=

81∗63

64∗105∗625∗63

104∗6=¿

81∗6105∗625∗104

= 486109∗625

= 486625000000000

=7,776∗10−10

3) 15-

log3( 2735 )+ log327−1

log30,3−2+log30.03− (log30,0003 )−4 log310

+ln e−3∗ln e−2=¿

log327243

+ log3127

log3( 310 )−2

+ log3( 3100 )−log3( 310000 )−4 log310

+ ln e−3 ln e−2=¿

−2−3

log31009

+log33100

−log33

10000−4 log310

4−3 ln e∗ln e−2=¿

−5

log3( 1009 ∗3

100 )−log3 310000

−log310000

−3∗1∗ln (e−2)=¿¿

−5

log3(1009

∗3

1003

10000)−log310000

−3∗(−2 ln e )=¿

−5

log3

1009

∗3

100∗100000

3 ∗1

10000

+6= −5

log319

+6=−5−2

+6=¿

5+122

=172