Documentation/FR/Math/Solutions entrainements
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Contents
1
Solutions des entraînements
1.1
Entraînement 1 (les bases)
1.2
Entraînement 2 (les parenthèses)
1.3
Entraînement 3 (les modèles)
1.4
Entraînement 4 (les symboles additionnels)
1.5
Entraînement 5 (les polices)
1.6
Entraînement 6 (les formules chimiques)
Solutions des entraînements
Entraînement 1
(les bases)
(
a
−
b
)
2
=
a
2
+
b
2
−
2
a
b
{\displaystyle {\left(a\mathrm {-} b\right)}^{2}\mathrm {=} {a}^{2}+{b}^{2}\mathrm {-} 2ab}
1
R
=
1
R
1
+
1
R
2
{\displaystyle {\frac {1}{R}}\mathrm {=} {\frac {1}{{R}_{1}}}+{\frac {1}{{R}_{2}}}}
3
m
⋅
5
m
=
15
m
2
{\displaystyle 3{\text{m}}\mathrm {\cdot } 5{\text{m}}\mathrm {=} 15{\text{m}}^{2}}
1
,
6726
⋅
10
−
27
kg
=
1
,
0073
u
{\displaystyle \mathrm {1,6726} \mathrm {\cdot } {10}^{\mathrm {-} 27}{\text{kg}}\mathrm {=} \mathrm {1,0073} {\text{u}}}
Entraînement 2
(les parenthèses)
a
⋅
b
c
=
a
⋅
b
c
{\displaystyle a\mathrm {\cdot } {\frac {b}{c}}\mathrm {=} {\frac {a\mathrm {\cdot } b}{c}}}
8
y
−
3
5
n
−
y
+
2
2
n
{\displaystyle {\frac {8y\mathrm {-} 3}{5n}}\mathrm {-} {\frac {y+2}{2n}}}
I
=
[
3
;
8
)
{\displaystyle I\mathrm {=} \mathrm {\lbrack } 3;8)\,}
{
x
=
2
{\displaystyle \mathrm {\lbrace } x\mathrm {=} 2\,}
Entraînement 3
(les modèles)
2
x
−
1
≤
3
−
5
x
{\displaystyle 2x\mathrm {-} 1\mathrm {\leq } 3\mathrm {-} 5x}
x
∈
A
{\displaystyle x\mathrm {\in } A}
,
A
∩
B
=
∅
{\displaystyle A\mathrm {\cap } B\mathrm {=} \mathrm {\varnothing } }
x
⋅
y
=
0
⇔
x
=
0
∨
y
=
0
{\displaystyle x\mathrm {\cdot } y\mathrm {=} 0\mathrm {\Leftrightarrow } x\mathrm {=} 0\mathrm {\vee } y\mathrm {=} 0}
x
2
3
=
x
2
/
3
{\displaystyle {\sqrt[{3}]{{x}^{2}}}\mathrm {=} {x}^{2\mathrm {/} 3}}
1
2
(
2
3
+
4
5
)
{\displaystyle {\frac {1}{2}}\left({\frac {2}{3}}+{\frac {4}{5}}\right)}
A
=
(
1
2
,
1
3
,
1
4
,
1
5
,
…
)
{\displaystyle A\mathrm {=} \left({\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},{\frac {1}{5}},\dots \right)}
∫
10
1
n
2
d
n
{\displaystyle {\underset {1}{\overset {10}{\mathrm {\int } }}}{n}^{2}{\mathit {dn}}}
2
(
x
−
3
)
+
2
=
4
(
x
+
2
)
⇔
2
x
−
4
=
4
x
+
8
⇔
x
=
−
6
{\displaystyle {\begin{array}{c}2\left(x\mathrm {-} 3\right)+2\mathrm {=} 4\left(x+2\right)\mathrm {\Leftrightarrow } \\2x\mathrm {-} 4\mathrm {=} 4x+8\mathrm {\Leftrightarrow } \\x\mathrm {=} \mathrm {-} 6\end{array}}}
Entraînement 4
(les symboles additionnels)
O
=
2
π
r
{\displaystyle O\mathrm {=} 2\pi r\,}
8
V
=
2
A
⋅
4
Ω
{\displaystyle 8{\text{V}}\mathrm {=} 2{\text{A}}\mathrm {\cdot } 4\Omega }
Q
Δ
t
{\displaystyle {\frac {Q}{\Delta t}}}
Entraînement 5
(les polices)
Entraînement 6
(les formules chimiques)
F
e
2
+
{\displaystyle {\mathrm {Fe} }^{2+}\,}
6
12
C
{\displaystyle {}_{6}^{12}\mathrm {C} \,}
C
H
4
+
2
O
2
→
C
O
2
+
2
H
2
O
{\displaystyle {\mathrm {CH} }_{4}+{\mathrm {2O} }_{2}\rightarrow {\mathrm {CO} }_{2}+2{\mathrm {H} }_{2}\mathrm {O} }
Chapitre précédent :
Annexe 5 Formules chimiques
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