DST2 corrigé

Corrigé du devoir surveillé 2
4(4 −4x)−2(7 −6x)=4×4−4×4x−(2 ×7−2×6x)
= 16 −16x−(14 −12x)
= 16 −16x−14 + 12x
= 2 −4x
(3 −x)2= 32−2×3×x+x2
= 9 −6x+x2
(6x−5)2= (6x)2−2×6x×5+52
= 36x2−60x+ 25
56x+ 35x2=7×8×x+7×5×x×x
=7x(8+5x)
9x2−121 = (3x)2−112
= (3x+11)(3x−11)
E
xE=xB+xD
2
=2+(−2)
2
= 0
yE=yB+yD
2
=0+4
2
= 2
E E(0; 2)
AB BC
AB = 7 BC = 4
AD
AD =p(xD−xA)2+ (yD−yA)2
=p((−2) −(−5))2+ (4 −0)2
=p(3)2+ (4)2=√9 + 16 = √25 = 5 AD = 5
−5−4−3−2−1 1 2
−1
1
2
3
4
5
A B
CD
EF
F[AC]F A C
yF=yA+yC
2=0+4 2=2=yE
E F

E
xE=xA+xC
2
=(−1) + 3
2
= 1
yE=yA+yC
2
=1,5+0
2
= 0,75
E E(1; 0,75)
BC =p(3 −2)2+ (0 −3)2
=p(1)2+ (−3)2
=√1 + 9 = √10
BC =√10
EB
EB =p(2 −1)2+ (3 −0,75)2
=p(1)2+ (2,25)2
=p1+5,0625 = p6,0625
•EC2+EB2= √73
4!2
+ (√6,0625)2= 10,625
•BC2= (√10)2= 10
10,625 6= 10 CEB
F[BD]
xF=xB+xD
2
=2+0
2
= 1
yF=yB+yD
2
=3+(−1,5)
2
= 0,75
F E E =F[AC] [BD]
ABCD
ABCD
CEB E
ABCD

Figure de l’exercice 3
−1.5−1−0.5 0.511.522.533.5
−2
−1.5
−1
−0.5
0.5
1
1.5
2
2.5
3
3.5
A
B
C
E
\
CEB = 86.59
D