Mqt1001 Question
Documents Gratuits : Mqt1001 Question. Recherche parmi 300 000+ dissertationsPar dan01 • 1 Octobre 2014 • 1 369 Mots (6 Pages) • 1 897 Vues
Question 1
A) (x + 25 / x+ 5) = (2x + 75 / 2x – 15)
Produits croisé
(x + 25) (2x -15) = (x + 5) (2x +75)
(x × 2x) (x × -15) (25 × 2x) (25 × -15) = (x × 2x) (x × 75) (5 × 2x) (5 × 75)
2x2 + -15x + 50x – 375 = 2x2 + 75x + 10x + 375
(2x2 - 2x2) (35x – 85x) (-375 – 375) = 0
X2 – 50x -750 = 0
Validation
x = -b ± √b2 4ac / 2a
X2 – 50x -750 = 0
X = (-50) + √ (-50)2 – 4 x 1 x (-750) / 2 x 1
X = 50 + √2 500 – (-3 000) / 2
X = 50 + 74, 16198 / 2 ou en moins X = 50 – 74, 16198 / 2
X = 124, 16198 / 2 x = -24, 16198 / 2
X = 62, 08 x = -12, 08
X2 – 50x -750 = 0
62,082 – 50(62, 08) – 750 = 0
3 854 - 3 104 – 750 = 0
X2 – 50x -750 = 0
-12, 082 – 50(-12, 08) – 750 = 0
146 + 604 – 750 = 0
B) (x+3 / 2x – 7) – (2x – 1 / x – 3) = 0
Produits croisés
(x + 3) (x – 3) = (2x – 7) (2x -1)
(x × x) (x × -3) (3 ×x) (3× -3) = (2x × 2x) (2x × -1) (-7 × 2x) (-7 × -1)
X2 -3x + 3x – 9 = 4x2 – 2x – 14x + 7
X2 – 9 = 4x2 – 16x + 7
(x2 – 4x2) + 16x (-9 -7) = 0
-3x2 + 16x – 16 = 0
Validation :
Formule : x = -b ± √b2 4ac / 2a
X = (-16) + √ (-16)2 – 4 x -3 x (-16) / 2 x -3
X = -16 + √256 – 192 / -6
X = -16 + √64 / -6 ou en moins…. X = -16 - √256 – 192 / -6
X = -16 + 8 / -6 x = -16 - 8 / -6
X = -8/-6 x = -24/ -6 = 4
-3x2 + 16x – 16 = 0
-3(4)2 + 16(4) – 16 = 0
-48 + 64 -16 = 0
-3x2 + 16x – 16 = 0
-3(-8/-6)2 + 16(-8/-6) – 16 = 0
-5,33333 + 21,33333 – 16 = 0
C) 14x – 4 / 3 ˂ 1 + 2x
3 (14x -4) / 3 ˂ 3 (1 + 2x)
14x – 4 ˂ 3 + 6x
(14x – 6x – 4 – 3) ˂ 0
8x – 7 ˂ 0
8x ˂ 7
8x /8 ˂ 7/8
X ˂ 0, 875
Tous les nombres plus petits que 0, 875
Question 2
A)
21/x + 12/y = 5 et 1/y – 1/x = 1/42
21/x + 12/y = 5 et (21) 1/y – 1/x = 1/42 (21)
21/x + 12/y = 5 et 21/y – 21/x = 21/42
21/x + 12/y = 5
+
21/y – 21/x = 21/42
12/y + 21/y = 33/y
33/y = 33/y divisé par 5.50
Y = 6
21/x + 12/6 = 5 (21/x + 12/6 – 12/6 = 5 – 12/6)
21/x = 3 divisé par 3 = X = 7
Validation :
21/7 + 12/6 = 5
3 + 2 = 5
Et
1/6 – 1/7 = 1/42
PPCM = 42
(7) 1/6 – (6) 1/7 = 1/42
7/42 – 6/42 = 1/42
B)1/(4x)+1/(3y)=2 et 1/y-1/(2x)=1
On multiplie la deuxième équation par 1/2
1/(2y)-1/(4x)=1/2
1/(4x)+1/(3y)=2
+
1/(2y)-1/(4x)=1/2
1/(3y)+1/(2y)=5/2
(5/6y)=5/2
(On multiplie les deux membres par l'inverse de 5/6 (donc 6/5)
on obtient 1/y=3
y=1/3
On prend la deuxième équation
1X3-1/(2x)=1
3-1/(2x)=1
3 -1/(2x) -3= (-3+1) = -2 donc 1/(2x)=2, on multiplie par l'inverse = 2
On trouve 1/x=4
x=1/4
Validation :
1/(4x)+1/(3y)=2 et 1/y-1/(2x)=1
1/ 4(1/4) + 1/ 3(1/3) = 2 et 1/ (1/3) – 1/ 2(1/4) = 1
1/1 + 1/1 = 2 et 3 – 2 = 1
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