() 3 –2 –3 6 –2 6 () Sol 9.pdf · QUELQUES RÉPONSES AUX EXERCICES 2C – JtJ 2018 Exercice...
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QUELQUES RÉPONSES AUX EXERCICES 2C – JtJ 2018 Quelques réponses : Thème 9 Exercice 9.1: a) 2 2 ( ) 3 = 2 2⋅3 = 2 6 = 64 b) 2 2 3 ( ) = 2 8 = 256 c) 2 3 ( ) 2 = 2 3⋅2 = 2 6 = 64 d) 2 3 – 3 2 = 8 – 9 = -1 e) 3 2 + 3 4 = 9 + 81 = 90 f) 10 3 + 10 2 = 1000 + 100 = 1100 g) 1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 = 1 4 3 4 = 1 81 h) − 2 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = −2 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = (−2) 3 5 3 = -8 125 i) − 3 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 = −3 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 = (−3) 4 2 4 = 81 16 j) 2 ( ) 4 = 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 2 = 4 k) 5 ( ) 6 =…= 5 3 = 125 l) − 1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 8 = −1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 8 =…= 1 3 4 = 1 81 Exercice 9.2: a) 2 6 2 2 = 2 6−2 = 2 4 = 16 b) −2 6 2 3 = − 2 6 2 3 = −2 6−3 = −2 3 = – 8 c) (−2) 6 2 3 = 2 6 2 3 = 2 6−3 = 2 3 = 8 d) 2 10 2 12 = 1 2 12−10 = 1 2 2 = 1 4 e) 2 4 −2 5 = − 2 4 2 5 = − 1 2 5−4 = – 1 2 f) 2 4 (−2) 5 = − 2 4 2 5 = − 1 2 5−4 = – 1 2 g) 2 2 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1 2 2 4 = 1 16 h) − 2 7 2 6 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = −2 7−6 ( ) 3 = (−2) 3 = – 8 i) 2 0 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1 2 4 2 = 1 16 j) 2 16 4 7 = 2 16 2 2 ( ) 7 = 2 16 2 14 = 2 16−14 = 2 2 = 4 k) 9 5 3 10 = 3 2 ( ) 5 3 10 = 3 10 3 10 = 1 j) 5 3 25 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 5 3 5 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1 25 Exercice 9.3: a) 2 -1 · 2 -2 = 2 -1 – 2 = 2 -3 = 1 8 b) 3 -4 · 3 4 = 3 -4 + 4 = 3 0 = 1 c) 4 0 · 4 -5 · 4 3 = 4 -2 = 1 16 d) a -3 · a 4 = a -3 + 4 = a 1 = a e) 2 −3 3 2 = 1 2 3 3 2 = 1 2 3 ⋅ 1 3 2 = 1 8 ⋅ 1 9 = 1 72 f) 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −2 = 1 (1 / 2) 2 = 1 1/4 = 1⋅ 4 = 4 g) 2 3 3 −2 = 8 1 3 2 = 8 ⋅ 3 2 = 72 h) − 2 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −2 = −2 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −2 = 5 −2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 5 2 (−2) 2 = 25 4 i) 1 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 · 2 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −2 = 1 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 · 5 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1 2 2 = 1 4 QUELQUES RÉPONSES AUX EXERCICES 2EC – JtJ 2018 Exercice 9.4: a) 2 3 · 2 4 · 2 -5 = 2 3 + 4 – 5 = 2 2 b) 3 4 ( ) 2 = 3 8 = 1 3 –8 c) 3 8 3 2 = 3 6 = 3 2 ( ) 3 = 9 3 d) 3 2 3 6 = 3 −4 = 3 2 ( ) −2 = 9 –2 e) 5 25 2 = 5 5 4 = 5 –3 f) 2 3 ⋅ 2 9 = 2 12 = 2 2 ( ) 6 = 4 6 g) 36 2 ⋅ 6 −2 = 6 4 ⋅ 6 −2 = 6 2 = 1 6 –2 h) − 1 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −3 = −1 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −3 = 5 −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = –5 ( ) 3 i) 1 25 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −3 = 25 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = 5 2 ( ) 3 = 5 6 j) 2 3 2 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −3 = 2 5 2 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = 2 2 ( ) 3 = 2 6 k) 2 −1 2 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 1 2 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 2 −4 ( ) 2 = 2 –8 l) 3 −4 9 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = 3 −8 9 4 = 3 2 ( ) −4 9 4 = 9 −4 9 4 = 9 –8 Exercice 9.5: Dans les 2 cas, il suffit de montrer que les valeurs des 2 côtés du "=" sont bien identiques. Exercice 9.6: a) 0 = 0 car 0 2 = 0 b) 0, 04 = 0, 2 car 0, 2 2 = 0, 04 c) 0, 0009 = 0, 03 car 0, 03 2 =… d) 1000 3 = 10 car 10 3 = 1000 e) 32 5 = 2 car 2 5 = 32 f) 16 4 = 2 car 2 4 = 16 g) 0, 027 3 = 0, 3 car 0, 3 3 =… h) 0, 0001 4 = 0,1 car 0,1 4 = 0, 0001 i) 0,125 3 = 0, 5 car 0, 5 3 =… j) 0, 0001 = 0, 01 car 0, 01 2 = 0, 0001 k) 0, 000008 3 = 0, 02 car 0, 02 3 = 0, 000008 Exercice 9.7: a) −27 3 = −3 car −3 ( ) 3 = −27 b) −1 5 = −1 car −1 ( ) 5 = −1 c) −4 2 est non définie d) (−2) 3 3 = −2 e) (−2) 2 = 2 car (−2) 2 = 4 f) −0,125 3 = −0, 5 car (−0, 5) 3 = −0,125 g) −0, 027 3 = −0, 3 car … h) −1 4 ( ) 2 est non définie i) (−2) 4 = 4
Transcript of () 3 –2 –3 6 –2 6 () Sol 9.pdf · QUELQUES RÉPONSES AUX EXERCICES 2C – JtJ 2018 Exercice...
QUELQUES RÉPONSES AUX EXERCICES
2C – JtJ 2018
Quelques réponses : Thème 9
Exercice 9.1: a) 22( )3= 22⋅3 = 26 = 64 b) 2 23( ) = 28 = 256
c) 23( )2= 23⋅2 = 26 = 64 d) 23 – 32 = 8 – 9 = -1
e) 32 + 34 = 9 + 81 = 90 f) 103 + 102 = 1000 + 100 = 1100
g) 1
3
⎛
⎝⎜⎞
⎠⎟4
=14
34=181
h) −2
5
⎛
⎝⎜
⎞
⎠⎟3
=−25
⎛
⎝⎜
⎞
⎠⎟3
=(−2)3
53=-8125
i) −3
2
⎛
⎝⎜
⎞
⎠⎟4
=−32
⎛
⎝⎜
⎞
⎠⎟4
=(−3)4
24=8116
j) 2( )4
= 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅2 = 4
k) 5( )6
=…= 53 =125 l) −1
3
⎛
⎝⎜
⎞
⎠⎟8
=−1
3
⎛
⎝⎜
⎞
⎠⎟8
=…=1
34=181
Exercice 9.2: a) 26
22= 26−2 = 24 =16 b)
−26
23= −
26
23= −26−3 = −23 = –8
c) (−2)6
23=26
23= 26−3 = 23 = 8 d)
210
212=
1
212−10=1
22=14
e) 24
−25= −
24
25= −
1
25−4= –12
f) 24
(−2)5= −
24
25= −
1
25−4= –12
g) 2
23⎛
⎝⎜
⎞
⎠⎟2
=1
22⎛
⎝⎜
⎞
⎠⎟2
=12
24=116
h) −27
26⎛
⎝⎜
⎞
⎠⎟
3
= −27−6( )3= (−2)3 = – 8
i) 20
22⎛
⎝⎜
⎞
⎠⎟
2
=1
4
⎛
⎝⎜⎞
⎠⎟2
=12
42=116
j) 216
47=216
22( )7 =
216
214= 216−14 = 22 = 4
k) 95
310=32( )
5
310=310
310=1 j)
53
252⎛
⎝⎜
⎞
⎠⎟
2
=53
54⎛
⎝⎜
⎞
⎠⎟
2
=1
5
⎛
⎝⎜⎞
⎠⎟2
=125
Exercice 9.3: a) 2-1 · 2-2 = 2-1 – 2 = 2-3 = 18
b) 3-4 · 34 = 3-4 + 4 = 30 = 1
c) 40 · 4-5 · 43 = 4-2 = 116
d) a-3 · a4 = a-3 + 4 = a1 = a
e) 2−3
32=
1
23
32=1
23⋅1
32=1
8⋅1
9=172
f) 1
2
⎛
⎝⎜⎞
⎠⎟−2
=1
(1/ 2)2=1
1/ 4=1⋅ 4 = 4
g) 23
3−2=81
32
= 8 ⋅32 = 72 h) −2
5
⎛
⎝⎜
⎞
⎠⎟−2
=−25
⎛
⎝⎜
⎞
⎠⎟−2
=5
−2
⎛
⎝⎜
⎞
⎠⎟2
=52
(−2)2=254
i) 1
5
⎛
⎝⎜⎞
⎠⎟2
·2
5
⎛
⎝⎜⎞
⎠⎟−2
=1
5
⎛
⎝⎜⎞
⎠⎟2
·5
2
⎛
⎝⎜⎞
⎠⎟2
=1
22=14
QUELQUES RÉPONSES AUX EXERCICES
2EC – JtJ 2018
Exercice 9.4: a) 23 · 24 · 2-5 = 23 + 4 – 5 = 22 b) 34( )2= 38 =
1
3–8
c) 38
32= 36 = 32( )
3= 93 d)
32
36= 3−4 = 32( )
−2= 9–2
e) 5
252=5
54= 5–3 f) 23 ⋅29 = 212 = 22( )
6= 46
g) 362 ⋅6−2 = 64 ⋅6−2 = 62 =1
6–2 h) −
1
5
⎛
⎝⎜
⎞
⎠⎟−3
=−15
⎛
⎝⎜
⎞
⎠⎟−3
=5
−1
⎛
⎝⎜
⎞
⎠⎟3
= –5( )3
i) 1
25
⎛
⎝⎜
⎞
⎠⎟−3
=25
1
⎛
⎝⎜
⎞
⎠⎟3
= 52( )3= 56 j)
23
25⎛
⎝⎜
⎞
⎠⎟
−3
=25
23⎛
⎝⎜
⎞
⎠⎟
3
= 22( )3= 26
k) 2−1
23⎛
⎝⎜
⎞
⎠⎟
2
=1
24⎛
⎝⎜
⎞
⎠⎟2
= 2−4( )2= 2–8 l)
3−4
92⎛
⎝⎜
⎞
⎠⎟
2
=3−8
94=32( )
−4
94=9−4
94= 9–8
Exercice 9.5: Dans les 2 cas, il suffit de montrer que les valeurs des 2 côtés du "=" sont bien identiques.
Exercice 9.6: a) 0 = 0 car 02 = 0 b) 0,04 = 0,2 car 0,22 = 0,04
c) 0,0009 = 0,03 car 0,032 =… d) 10003 =10 car 103 =1000
e) 325 = 2 car 25 = 32 f) 164 = 2 car 24 =16
g) 0,0273 = 0,3 car 0,33 =… h) 0,00014 = 0,1 car 0,14 = 0,0001
i) 0,1253 = 0,5 car 0,53 =… j) 0,0001 = 0,01 car 0,012 = 0,0001
k) 0,0000083 = 0,02 car 0,023 = 0,000008
Exercice 9.7: a) −273 = −3 car −3( )3 = −27 b) −15 = −1 car −1( )5 = −1
c) −42 est non définie d) (−2)33 = −2
e) (−2)2 = 2 car (−2)2 = 4 f) −0,1253 = −0,5 car (−0,5)3 = −0,125
g) −0,0273 = −0,3 car … h) −14( )2
est non définie
i) (−2)4 = 4
QUELQUES RÉPONSES AUX EXERCICES
2C – JtJ 2018
Exercice 9.8: a) 1
43 ⋅
1
23 =
1
83 =
12
b) 22 = 2
c) 9+1 = 10 d) 84 ⋅ 324 = 2564 = 4
e) 26 = 64 = 8 f) 9 + 3 = 3 + 3
g) −27
10003 =
−273
10003=–310
h) 0,00014( )4
= 0, 0001
i) 2105 = 22( )5
5 = 22 = 4
Exercice 9.9: a) 2700 = 100 ⋅ 27 ≅10 ⋅5,19 ≅ 51,9
b) 27'000 = 100 ⋅ 270 ≅10 ⋅16, 43≅164,3
c) 270'000 = 1000 ⋅ 27 ≅100 ⋅5,19 = 519
d) 0,27 =27
100=
27
10≅ 0,519
e) 0,027 =270
10000=
270
100= 0,1643
f) 0,000027 =27
1000000=
27
1000≅5,19
1000≅ 0,00519
Exercice 9.10: a) 27'0003 = 10003 ⋅ 273 =10 ⋅3= 30
b) 270'0003 = 10003 ⋅ 2703 ≅10 ⋅6, 46 = 64,6
c) 2'700'0003 = 10003 ⋅ 27003 ≅10 ⋅13,92 =139,2
d) 0,273 =270
10003 =
2703
10≅6, 46
10≅ 0,646
e) 0,0273 =27
10003 =
273
10=3
10= 0,3
f) 0,000273 =270
10000003 =
2703
100≅6, 46
100= 0,0646
Exercice 9.11: a) 300 ≅17,3
b) 3
3≅1, 73
c) 4
3=2
3=2 ⋅ 33
≅3, 46
3≅1,15
QUELQUES RÉPONSES AUX EXERCICES
2EC – JtJ 2018
Exercice 9.12: a) 10 3 b) 3 c) 2 3
3 d) 23 2 = 8 2
e) 20 10 f) 26
2 g) 5 3 h) 2+ 2
i) 3 3
2 j)
2 3 +33
k) 9 2 l) 2
2
Exercice 9.13: a) 5 b) 3 c) 9
Exercice 9.14: a) 51/2 = 5 b) 43/2 = 41/2( )3= 4( )
3
= 8
c) 161/4 = 24( )1/4= 2 d) 321/10 = 321/5( )
1/2= 21/2 = 2
e) 93/2 = 91/2( )3= 33 = 27 f) 250,5 = 251/2 = 5
g) 1
9
⎛
⎝⎜⎞
⎠⎟−12
= 91/2 = 3 h) 27
8
⎛
⎝⎜
⎞
⎠⎟
23
=27
83⎛
⎝⎜
⎞
⎠⎟
2
=3
2
⎛
⎝⎜⎞
⎠⎟2
=94
i) 2−1/2 =1
21/2=1
2=22 j) 9
−1,5 =1
93/2=
1
91/2( )3 =
127
k) 56/5 = 56( )
1/5
= 565 = 5 55 l) 2
81/2=2
8=
2
2 2=22
Exercice 9.15: a) 31/5 b) 56/11 c) 33/9 = 31/3
d) 2-1/2 e) 3
2
⎛
⎝⎜⎞
⎠⎟1/2
f) 39/3 = 33
Exercice 9.16: a) 24 ⋅ 2 = 21/4 ⋅21/2 = 21/4+1/2 = 23/4 = 234 = 84
b) 34 ⋅ 323 = 31/4 ⋅32/3 = 31/4+2/3 = 311/12 = 31112
c) 253 ⋅ 56 = 523 ⋅ 56 = 52/3 ⋅51/6 = 52/3+1/6 = 55/6 = 556
d) 3
33=31/2
31/3= 31/2−1/3 = 31/6 = 36
e) 2
224=21/2
22/4=1
f) 2 ⋅ 2 ⋅ 2 = 2 ⋅ 2 ⋅21/2( )1/2( )
1/2
= 21/2 ⋅21/4 ⋅21/8 = 21/2+1/4+1/8 = 27/8 = 278
QUELQUES RÉPONSES AUX EXERCICES
2C – JtJ 2018
Exercice 9.17: Il s'agit de proposer une chaîne d'égalités passant du terme de gauche à celui
de droite. Mais… Ne regardez pas trop vite ci-dessous…
a) amn = a1/mn = a1/m( )1/n= a
1m ⋅1n = a
1n ⋅m = an⋅m
b) am⋅pm⋅n = am⋅p( )1m⋅n = a
m⋅pm⋅n = a
pn = ap( )
1n = apn
c) 26 d) 232 = 2 2 e) 6
Exercice 9.18: Vous avez dû obtenir un nombre proche de 2. Voici pourquoi il s'agit même
exactement de 2.
1) Partons de 2 = 4 que l'on peut écrire: 2 = 2+ 2
2) Substituons le dernier 2 grâce à cette même dernière égalité. On obtient:
2 = 2+ 2+ 2
3) Réitérerons ce même procédé:
2 = 2+ 2+ 2+ 2
4) Et encore une fois:
2 = 2+ 2+ 2+ 2+ 2
5) Et ainsi de suite:
2 = 2+ 2+ 2+ 2+…
Je vous laisse trouver l'imbrication de racines qui donne 3 à la place de 2.
Exercice 9.19: Exercice BONUS.
