1 Simplification 2 Résolution d`équations 3 Limites, dérivées

4
√81; 3
√a5∗3 √a7;e1.2∗e;e2
√e; (ex+e−x)2−(ex−e−x)2;e−ln 1
3; ln(√e) + ln(1
e)
ln 2 ln 5
ln 1
2+ ln 2 3+··· + ln 98 99 + ln 99 100
A= ln(√e3+eln√3); B= ln 4
√e;C=e−2 ln 2;D=eln 3−ln 2
eln 3+ln 2
e3x−1= 1; (ex−e−x)2= 0; 2x+3 = 3x+2;5
√x= 2x; 2x−1+ 22−x= 3
ln(x+ 1) + ln(x+ 3) = ln(x+ 7)
lim
x→e
ln x−1
x−e; lim
x→0
ex2−1
x; lim
x→0
ln(1 + 5x)
x; lim
x→(1
e)+
ln x−2
ln x+ 1; lim
x→+∞(x−ln x); lim x→+∞
e2x
ex+x2;
ex2+5x;e−2e−5x; ln(ex−3); ln(x−1
x+ 1); xln x−x;xx; (3x+ 2)√3x+ 2; (x2+ 1)0,2
f
f(x) = ex−1
ex+ 1
f
f IR
f−∞ limx→+∞f(x)
f

f IR ]−1,1[
y∈]−1,1[ y=f(x)x y
g f
g
CfCgO(0,0)
f:t7→ x0
1 + ea−bt
x0a b x0= 10 a= 2 b= 0.5
f
f
Ω(4,5)
f
f0=1
20f(10 −f)
4
√81 = (81)1
4= (92)1 4= ((32)2)1 4= 32∗2 4= 3
3
√a5∗3 √a7=a5
3∗a7 3=a5 3+7 3=a4
e1.2∗e=e1.2∗e1=e1.2+1 =e2.2
e2
√e=e2
e1
2
=e2∗e−1
2=e2−1 2=e3 2=e∗√e
(ex+e−x)2−(ex−e−x)2= 2ex∗2e−x= 4e0= 4
e−ln 1
3=eln(3) = 3
ln(√e) + ln(1
e) = 1 2−1 = −1 2

ln 1
2+ ln 2 3+··· + ln 98 99 + ln 99 100 = ln(1 2∗2 3∗ ··· ∗ 98 99 ∗99 100) = ln( 1 100) = −ln(100)
100 = 102= (5 ∗2)2
ln 1
2+ ln 2 3+··· + ln 98 99 + ln 99 100 =−ln((5 ∗2)2) = −2 ln(5 ∗2) = −2(ln 5 + ln 2)
log 1
2+ log 2 3+··· + log 98 99 + log 99 100 =−log(100) = −log(102) = −2
A B =C=D=1
4
e3x−1= 1 ⇔e3x−1=e0⇔3x−1 = 0 ⇔x=1
3
S={1
3}
(ex−e−x)2= 0 ⇔ex−e−x= 0 ⇔ex=e−x⇔x=−x⇔x= 0
S={0}
2x+3 = 3x+2 ⇔2x∗23= 3x∗32⇔2x
3x=32 23⇔µ2 3¶x
=9
8⇔exln( 2
3)=eln( 9 8)⇔xln(2
3) = ln(9 8)⇔x=ln(9
8)
ln(2
3)
S={ln( 9
8)
ln( 2
3)}
5
√x= 2x0
x6= 0
5
√x= 2x⇔x1
5
x= 2 ⇔x1
5−1= 2 ⇔x−4 5= 2 ⇔³x−4 5´5= 25⇔x−4= 25⇔x4= 2−5
x7→ x4
x4=1
25⇔x=µ1 25¶1
4
x=−µ1
25¶1
4
S={0,1
24
√2,−1
24
√2}
2x−1+ 22−x= 3 ⇔2x
2+4 2x= 3 ⇔½X
2+4 X−3 = 0
X= 2x>0⇔½X2
2+ 4 −3X= 0
X= 2x>0
∆ = 1 >0X= 4 >0X= 2 >0 2x= 4 = 222x= 2 = 21
S={1,2}
x+ 1 >0x+ 3 >0x+ 7 >0x∈]−1,+∞[
x > −1
ln(x+1)+ln(x+3) = ln(x+7) ⇔ln((x+1)(x+3)) = ln(x+7) ⇔(x+1)(x+3) = x+7 ⇔x2+3x−4 = 0
∆ = 25 >0x= 1 >−1x=−4<−1
S={1}

lim
x→e
ln x−1
x−e= lim
x→e
ln x−ln e
x−e=1 e
lim
x→0
ex2−1
x= lim
x→0
ex2−1
x2∗x= 0 lim
X→0
eX−1
X= 1
lim
x→0
ln(1 + 5x)
x= lim
x→0
ln(1 + 5x)
5x∗5 = 5 lim
X→0
ln(1 + X)
X= 1
lim
x→(1
e)+
ln x−2
ln x+ 1 =−∞ lim
x→(1
e)+(ln x−2) = −3 lim
x→(1
e)+(ln x+ 1) = 0+
lim
x→+∞(x−ln x) = lim x→+∞
x(1 −ln x
x) = +∞lim
x→+∞
ln x
x= 0
lim
x→+∞
e2x
ex+x2= lim
x→+∞
1
e−x+x2e−2x= +∞lim
x→+∞
x2e−2x= 0+
x∈IR, (ex2+5x)0= (2x+ 5)ex2+5x
x∈IR, (e−2e−5x)0= (−2e−5x)0e−2e−5x= 10e−5xe−2e−5x= 10e−2e−5x−5x
x > ln 3,(ln(ex−3))0=ex
ex−3
x∈]− ∞,−1[∪]1,+∞[,(ln(x−1
x+ 1))0=
(x+1)−(x−1)
(x+1)2
x−1
x+1
=2
(x−1)(x+ 1)
x∈IR∗
+,(xln x−x)0= (ln x+ 1) −1 = ln xln
x∈IR∗
+,(xx)0= (exln x)0= (xln x)0exln x= (ln x+ 1)xx
x∈[−2
3,+∞[,((3x+ 2)√3x+ 2)0= ((3x+ 2)3
2)0=3
2∗3∗(3x+ 2)1
2=9
2√3x+ 2
x∈IR, ((x2+ 1)0,2)0= 0.2∗2x∗(x2+ 1)−0,8=0.4x
(x2+ 1)0,8
f
f(x) = ex−1
ex+ 1
f
f x ex+ 1 6= 0 x ex+ 1 >1Df=IR
x∈ Df−x∈ Dff(−x) = e−x−1
e−x+1 =1−ex 1+ex=−f(x)
f f IR−IR+
limx→−∞ ex= 0 lim−∞ f=−1
lim+∞f= 1
f IR ∀x∈IR, f0(x) = ex(ex+1)−ex(ex−1)
(ex+1)2=2ex (ex+1)2>0f
IR
flim−∞ f=−1 lim+∞f= 1 f(0) = 0

f IR f IR
f(IR) =] −1,1[
y∈]−1,1[
y=ex−1
ex+ 1 ⇔y(ex+ 1) = (ex−1) ⇔y+ 1 = ex(1 −y)⇔1 + y 1−y=ex⇔ln(1 + y 1−y) = x
f x ∈]−1,1[ g(x) = ln(1+x
1−x)
x∈]−1,1[ 1+x
1−x>0
g g0(x) = 2
(1+x)(1−x)x∈]−1,1[
g]−1,1[ lim−1g=−∞ lim1g= +∞g(0) = 0
f g
CfCgy=x
Cfx=−1x= 1 −∞
+∞(0,0) y=f0(0)(x−0) + f(0) = 1
2x
Cg−1 1 (0,0) y= 2x
g0(0) = 1
f0(g(0))
f:t7→ 10
1 + e2−0.5t
Df=IR lim−∞ f= 0 lim+∞f= 10
f IR x ∈IR
f0(x) = 10 ∗(−1) (−0.5)e2−0.5t
(1 + e2−0.5t)2=5e2−0.5t (1 + e2−0.5t)2>0
f IR
f y = 0 y= 10
−∞ +∞(4,5)
Ω(4,5)
M(t, y) (0,−→ i , −→ j) (T, Y )
(Ω,−→ i , −→ j)−−→ OM =−→ OΩ + −−→ ΩM½t= 4 + T
y= 5 + Y
f(Ω,−→ i , −→ j)
ΩCf
t∈IR T ∈IR −T∈IR
M∈ Cf⇔y=f(t)⇔Y+ 5 = 10
1 + e2−0.5(T+4) ⇔Y=10
1 + e−0.5T−5⇔Y= 5(1−e−0.5T 1 + e−0.5T)
T7→ 5(1−e−0.5T
1+e−0.5T) Ω(4,5)
f
t
f0(t) = 5e2−0.5t
(1 + e2−0.5t)2