jaqueline j. lugo matemática de nivelación 2 b
TRANSCRIPT
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
