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En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Exposants rationnelsExpressions algébriques/Exercices/Exposants rationnels », n'a pu être restituée correctement ci-dessus.
Simplifier les expressions :
a)
(
x
1
2
−
x
1
4
+
1
)
(
x
1
2
+
x
1
4
+
1
)
(
x
−
x
1
2
+
1
)
{\displaystyle \left(x^{\frac {1}{2}}-x^{\frac {1}{4}}+1\right)\left(x^{\frac {1}{2}}+x^{\frac {1}{4}}+1\right)\left(x-x^{\frac {1}{2}}+1\right)}
b)
(
2
1
4
a
+
3
1
4
b
)
(
3
1
4
a
−
2
1
4
b
)
−
6
1
4
(
a
2
−
b
2
)
+
2
1
2
a
b
{\displaystyle \left(2^{\frac {1}{4}}a+3^{\frac {1}{4}}b\right)\left(3^{\frac {1}{4}}a-2^{\frac {1}{4}}b\right)-6^{\frac {1}{4}}\left(a^{2}-b^{2}\right)+2^{\frac {1}{2}}ab}
c)
x
−
y
x
3
4
+
x
1
2
y
1
4
.
x
1
2
y
1
4
+
x
1
4
y
1
2
x
1
2
+
y
1
2
{\displaystyle {\frac {x-y}{x^{\frac {3}{4}}+x^{\frac {1}{2}}y^{\frac {1}{4}}}}.{\frac {x^{\frac {1}{2}}y^{\frac {1}{4}}+x^{\frac {1}{4}}y^{\frac {1}{2}}}{x^{\frac {1}{2}}+y^{\frac {1}{2}}}}}
d)
x
1
2
−
2
x
1
4
+
1
x
1
4
−
2
x
1
8
+
1
{\displaystyle {\frac {x^{\frac {1}{2}}-2x^{\frac {1}{4}}+1}{x^{\frac {1}{4}}-2x^{\frac {1}{8}}+1}}}
a) Corrigé
On trouve :
x
2
+
x
+
1
{\displaystyle x^{2}+x+1}
b) Corrigé
On trouve :
3
1
2
a
b
{\displaystyle 3^{\frac {1}{2}}ab}
c) Corrigé
On trouve :
x
1
4
y
1
4
−
x
−
1
4
y
3
4
{\displaystyle x^{\frac {1}{4}}y^{\frac {1}{4}}-x^{\frac {-1}{4}}y^{\frac {3}{4}}}
d) Corrigé
On trouve :
(
x
1
8
+
1
)
2
{\displaystyle \left(x^{\frac {1}{8}}+1\right)^{2}}
Démontrer que le polynôme :
P
=
x
3
−
a
−
2
3
b
−
1
(
a
2
+
b
2
)
x
+
b
1
2
{\displaystyle P=x^{3}-a^{-{\frac {2}{3}}}b^{-1}\left(a^{2}+b^{2}\right)x+b^{\frac {1}{2}}}
s'annule si l'on remplace
x
{\displaystyle x}
a
2
3
b
−
1
2
{\displaystyle a^{\frac {2}{3}}b^{-{\frac {1}{2}}}}
Factoriser
P
{\displaystyle P}
Solution
(
a
2
3
b
−
1
2
)
3
−
a
−
2
3
b
−
1
(
a
2
+
b
2
)
(
a
2
3
b
−
1
2
)
+
b
1
2
=
a
2
b
−
3
2
−
(
a
2
+
b
2
)
b
−
3
2
+
b
1
2
=
0
{\displaystyle \left(a^{\frac {2}{3}}b^{-{\frac {1}{2}}}\right)^{3}-a^{-{\frac {2}{3}}}b^{-1}\left(a^{2}+b^{2}\right)\left(a^{\frac {2}{3}}b^{-{\frac {1}{2}}}\right)+b^{\frac {1}{2}}=a^{2}b^{-{\frac {3}{2}}}-\left(a^{2}+b^{2}\right)b^{-{\frac {3}{2}}}+b^{\frac {1}{2}}=0}
P
=
(
x
−
a
2
3
b
−
1
2
)
(
x
2
+
a
2
3
b
−
1
2
x
+
a
−
2
3
b
)
{\displaystyle P=\left(x-a^{\frac {2}{3}}b^{-{\frac {1}{2}}}\right)\left(x^{2}+a^{\frac {2}{3}}b^{-{\frac {1}{2}}}x+a^{-{\frac {2}{3}}}b\right)}
Vérifier l'identité :
(
a
2
+
a
4
3
b
2
3
)
1
2
+
(
b
2
+
a
2
3
b
4
3
)
1
2
=
(
a
2
3
+
b
2
3
)
3
2
{\displaystyle \left(a^{2}+a^{\frac {4}{3}}b^{\frac {2}{3}}\right)^{\frac {1}{2}}+\left(b^{2}+a^{\frac {2}{3}}b^{\frac {4}{3}}\right)^{\frac {1}{2}}=\left(a^{\frac {2}{3}}+b^{\frac {2}{3}}\right)^{\frac {3}{2}}}
Solution
Posons
x
=
a
2
3
{\displaystyle x=a^{\frac {2}{3}}}
y
=
b
2
3
{\displaystyle y=b^{\frac {2}{3}}}
(
x
3
+
x
2
y
)
1
2
+
(
y
3
+
x
y
2
)
1
2
=
(
x
2
)
1
2
(
x
+
y
)
1
2
+
(
y
2
)
1
2
(
y
+
x
)
1
2
=
(
x
+
y
)
3
2
{\displaystyle \left(x^{3}+x^{2}y\right)^{\frac {1}{2}}+\left(y^{3}+xy^{2}\right)^{\frac {1}{2}}=\left(x^{2}\right)^{\frac {1}{2}}\left(x+y\right)^{\frac {1}{2}}+\left(y^{2}\right)^{\frac {1}{2}}\left(y+x\right)^{\frac {1}{2}}=\left(x+y\right)^{\frac {3}{2}}}
En supposant
a
>
0
{\displaystyle a>0}
b
>
0
{\displaystyle b>0}
a
1
2
−
[
a
−
(
a
2
−
a
x
)
1
2
]
1
2
a
1
2
+
[
a
−
(
a
2
−
a
x
)
1
2
]
1
2
{\displaystyle {\frac {a^{\frac {1}{2}}-\left[a-(a^{2}-ax)^{\frac {1}{2}}\right]^{\frac {1}{2}}}{a^{\frac {1}{2}}+\left[a-(a^{2}-ax)^{\frac {1}{2}}\right]^{\frac {1}{2}}}}}
pour :
x
=
a
[
1
−
16
b
2
(
1
+
b
)
4
]
{\displaystyle x=a\left[1-{\frac {16b^{2}}{(1+b)^{4}}}\right]}
Corrigé
On trouve :
Si
b
<
1
{\displaystyle b<1}
b
{\displaystyle b}
Si
b
>
1
{\displaystyle b>1}
1
b
{\displaystyle {\frac {1}{b}}}
p
{\displaystyle p}
q
{\displaystyle q}
a
+
b
a
−
b
{\displaystyle {\frac {a+b}{a-b}}}
P
=
1
2
.
a
2
−
b
2
a
2
+
b
2
(
x
1
p
+
x
1
q
)
{\displaystyle P={\frac {1}{2}}.{\frac {a^{2}-b^{2}}{a^{2}+b^{2}}}\left(x^{\frac {1}{p}}+x^{\frac {1}{q}}\right)}
pour :
x
=
(
a
+
b
a
−
b
)
2
p
q
q
−
p
{\displaystyle x=\left({\frac {a+b}{a-b}}\right)^{\frac {2pq}{q-p}}}
Corrigé
On trouve :
(
a
+
b
a
−
b
)
q
+
p
q
−
p
{\displaystyle \left({\frac {a+b}{a-b}}\right)^{\frac {q+p}{q-p}}}
En supposant
p
>
0
,
q
>
0
,
p
q
>
1
,
{\displaystyle p>0,\,q>0,\,pq>1,}
(
p
2
−
1
q
2
)
p
(
p
−
1
q
)
q
−
p
(
q
2
−
1
p
2
)
q
(
q
+
1
p
)
p
−
q
{\displaystyle {\frac {\left(p^{2}-{\frac {1}{q^{2}}}\right)^{p}\left(p-{\frac {1}{q}}\right)^{q-p}}{\left(q^{2}-{\frac {1}{p^{2}}}\right)^{q}\left(q+{\frac {1}{p}}\right)^{p-q}}}}
Corrigé
On trouve :
(
p
q
)
p
+
q
{\displaystyle \left({\frac {p}{q}}\right)^{p+q}}
Simplifier les fractions :
a)
a
2
−
a
3
2
x
1
2
−
2
a
1
2
x
1
4
+
2
x
3
4
a
1
2
−
x
1
2
{\displaystyle {\frac {a^{2}-a^{\frac {3}{2}}x^{\frac {1}{2}}-2a^{\frac {1}{2}}x^{\frac {1}{4}}+2x^{\frac {3}{4}}}{a^{\frac {1}{2}}-x^{\frac {1}{2}}}}}
b)
x
5
3
−
x
4
3
y
1
3
−
x
y
2
3
+
x
2
3
y
x
5
3
−
2
x
y
2
3
+
x
1
3
y
4
3
{\displaystyle {\frac {x^{\frac {5}{3}}-x^{\frac {4}{3}}y^{\frac {1}{3}}-xy^{\frac {2}{3}}+x^{\frac {2}{3}}y}{x^{\frac {5}{3}}-2xy^{\frac {2}{3}}+x^{\frac {1}{3}}y^{\frac {4}{3}}}}}
c)
a
−
x
+
4
a
1
4
x
3
4
−
4
a
1
2
x
1
2
a
1
2
+
2
a
1
4
x
1
4
−
x
1
2
{\displaystyle {\frac {a-x+4a^{\frac {1}{4}}x^{\frac {3}{4}}-4a^{\frac {1}{2}}x^{\frac {1}{2}}}{a^{\frac {1}{2}}+2a^{\frac {1}{4}}x^{\frac {1}{4}}-x^{\frac {1}{2}}}}}
d)
x
2
3
+
2
x
1
3
−
16
x
−
2
3
−
32
x
−
1
x
1
6
+
4
x
−
1
6
+
4
x
−
1
2
{\displaystyle {\frac {x^{\frac {2}{3}}+2x^{\frac {1}{3}}-16x^{-{\frac {2}{3}}}-32x^{-1}}{x^{\frac {1}{6}}+4x^{-{\frac {1}{6}}}+4x^{-{\frac {1}{2}}}}}}
e)
1
+
x
−
3
2
x
1
3
(
1
+
x
1
2
)
+
x
1
2
(
2
+
9
16
x
1
6
)
1
−
3
4
x
1
3
+
x
1
2
{\displaystyle {\frac {1+x-{\frac {3}{2}}x^{\frac {1}{3}}\left(1+x^{\frac {1}{2}}\right)+x^{\frac {1}{2}}\left(2+{\frac {9}{16}}x^{\frac {1}{6}}\right)}{1-{\frac {3}{4}}x^{\frac {1}{3}}+x^{\frac {1}{2}}}}}
a) Corrigé
On trouve :
a
3
2
−
2
x
1
4
{\displaystyle a^{\frac {3}{2}}-2x^{\frac {1}{4}}}
b) Corrigé
On trouve :
x
1
3
x
1
3
+
y
1
3
{\displaystyle {\frac {x^{\frac {1}{3}}}{x^{\frac {1}{3}}+y^{\frac {1}{3}}}}}
c) Corrigé
On trouve :
(
x
1
4
−
a
1
4
)
2
{\displaystyle \left(x^{\frac {1}{4}}-a^{\frac {1}{4}}\right)^{2}}
d) Corrigé
On trouve :
(
x
1
3
−
2
)
(
x
2
3
+
4
)
x
−
1
2
{\displaystyle \left(x^{\frac {1}{3}}-2\right)\left(x^{\frac {2}{3}}+4\right)x^{-{\frac {1}{2}}}}
e) Corrigé
On trouve :
x
1
2
−
3
4
x
1
3
+
1
{\displaystyle x^{\frac {1}{2}}-{\frac {3}{4}}x^{\frac {1}{3}}+1}
![{\displaystyle {\frac {a^{\frac {1}{2}}-\left[a-(a^{2}-ax)^{\frac {1}{2}}\right]^{\frac {1}{2}}}{a^{\frac {1}{2}}+\left[a-(a^{2}-ax)^{\frac {1}{2}}\right]^{\frac {1}{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7286fee9e5662c376c98442ab5ccaf5cbec64186)
![{\displaystyle x=a\left[1-{\frac {16b^{2}}{(1+b)^{4}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c3ae9e8f031d12a368ade89658543965e96d644)
