Exercice 1 :
Je pose (x-1)(5+x) = 40 :
(x-1)(5+x)-40 = 0
5x+x²-5-x-40 = 0
x²+4x-45 = 0
Δ = b²-4ac
Δ = 4²-4*1*(-45)
Δ = 196
Δ > 0 donc l'équation admet deux solutions distinctes x1 et x2
x1 = (-b-√Δ)/2a
x1 = (-4-√196)/2*1
x1 = -9
x2 = (-b+√Δ)/2a
x2 = (-4+√196)/2*1
x2 = 5
Les solutions de l'équations sont donc x = -9 et x = 5.
Exercice 2 :
Pour que (x²-4)/((x-7)(6+x)) soit égale à 0 il faut que le nominateur soit nul :
Je résous donc l'équation x²-4 = 0 :
x²-4 = 0
(x-2)(x+2) = 0
soit
x-2 = 0
x = 2
ou
x+2 = 0
x = -2
Exercice 3 :
1) 4x+7 = -6x-4-2x
12x = -11
x = -(11/12)
2) (9/7)x = -(21/4)
x = -(21/4)/(9/7)
x = -(21/4)*(7/9)
x = -(147/36)
x = -(49/12)
3) 10x-10-6x = 6-5x
9x = 16
x = 16/9
4) 4x-8x+2 = -7-6x+12
2x = 3
x = 3/2
Exercice 4 :
1) 4-2x-7x = 5x-2
-14x = -6
x = 6/14
x = 3/7
2) (7/12)x-1 = (x/6)+(1/4)
(7/12)x-(1/6)x = (1/4)+1
(7/12)x-(2/12)x = (1/4)+(4/4)
(5/12)x = 5/4
x = (5/4)/(5/12)
x = (5/4)*(12/5)
x = 60/20
x = 3
3) -3(5x-2)-(2x+1) = 0
-15x+6-2x-1 = 0
-17x = -5
x = 5/17
4) -(5x+6)+4(x-3) = 0
-5x-6+4x-12 = 0
-x = 18
x = -18
5) (3x-5)² = (2+3x)(3x-1)
9x²-30x+25 = 6x-2+9x²-3x
-33x = 27
x = -(27/33)
x = -(9/11)
6) (3x/10)-(1/5) = 4-((x+6)/5)
(3x/10)+((x+6)/5) = 4+(1/5)
(3x/10)+((2x+12)/10) = (20/5)+(1/5)
(5x+12)/10 = 21/5
(5x/10)+(12/10) = 21/5
5x/10 = (21/5)-(12/10)
5x/10 = (42/10)-(12/10)
5x/10 = 30/10
5x = 30
x = 6
Exercice 5 :
Dans cet exercice je factorise les expressions
1) 9x²-25 = 0
(3x-5)(3x+5) = 0
soit
3x-5 = 0
3x = 5
x = 5/3
ou
3x+5 = 0
3x = -5
x = -(5/3)
2) 16x²-48x+36 = 0
(4x-6)² = 0
<=> (4x-6)(4x-6) = 0
4x-6 = 0
4x = 6
x = 6/4
x = 3/2
3) 16x²+24x+9
(4x+3)² = 0
<=> (4x+3)(4x+3) = 0
4x+3 = 0
4x = -3
x = -(3/4)
4) x²-41 = -5
x²-36 = 0
(x-6)(x+6) = 0
soit
x-6 = 0
x = 6
ou
x+6 = 0
x = -6
5) (x+10)² = 9
(x+10)²-9 = 0
((x+10)-3)((x+10)+3) = 0
(x+7)(x+13) = 0
soit
x+7 = 0
x = -7
ou
x+13 = 0
x = -13
6) x² = -81
x²+81 = 0
(x+9)(x+9) = 0
x+9 = 0
x = -9
Or, on sait que le carré d'un nombre est toujours positif donc cette équation n'admet pas de solution sur R.
Exercice 6 :
Dans cet exercice, je factorise les expressions
1) (1+4x)²-(5x-2)(1+4x) = 0
(1+4x)[(1+4x)-(5x-2)]
(1+4x)(-x+3) = 0
soit
1+4x = 0
4x = -1
x = -(1/4)
ou
-x+3 = 0
-x = -3
x = 3
2) (x+5)² = (9x+1)²
(x+5)²-(9x+1)² = 0
(((x+5)-(9x+1))((x+5)+(9x+1)) = 0
(-8x+4)(10x+6) = 0
soit
-8x+4 = 0
-8x = -4
x = 4/8
x = 1/2
ou
10x+6 = 0
10x = -6
x = -(6/10)
x = -(3/5)
3) 9-4x²-(3x-2)(5-x) = 0
(3-2x)(3+2x)-(3x-2)(5-x) = 0
(3-2x)[(3x-2)-(5-x)] = 0
(3-2x)(2x-7) = 0
soit
3-2x = 0
-2x = -3
x = 3/2
ou
2x-7 = 0
2x = 7
x = 7/2
Exercice 7 :
1) (x-1)/(x+6)-3 = 0
(x-1)/(x+6)-(3x-18)/(x+6) = 0
(-2x+19)/(x+6) = 0
-2x+19 = 0
-2x = -19
x = 19/2
2) x/(2-x)+x+3 = 0
x/(2-x)+(2x-x²)/(2-x)+(6-3x)/(2-x) = 0
(-x²-x+6)/(2-x) = 0
Δ = b²-4ac
Δ = (-1)²-4*(-1)*6
Δ = 25
Δ > 0 donc l'équation admet deux solutions distinctes x1 et x2
x1 = (-b-√Δ)/2a
x1 = (1-√25)/2*(-1)
x1 = 2
x2 = (-b+√Δ)/2a
x2 = (1+√25)/2*(-1)
x2 = -3
Les solutions de l'équations sont donc x = -3 et x = 2.
3) (4-x)/(7x-7) = 1
(4-x)/(7x-7)-1 = 0
(4-x)/(7x-7)-(7x-7)/(7x-7) = 0
(-8x+11)/(7x-7) =0
-8x+11 = 0
-8x = -11
x = 11/8
4) 5/(2-5x) = 4/(1-x)
5/(2-5x)-4/(1-x) = 0
(5-5x)/(2-5x²)-(8-20x)/(2-5x²) = 0
(15x-3)/(2-5x²) = 0
15x-3 = 0
15x = 3
x = 3/15
x = 1/5
