INECUACIONES DE SEGUNDO GRADO 1) Al resolver el sistema: 1 2 35 π₯ β 3π₯ + >0 2 8 (2x-7)(2x-5) X=1/2 X=5/2 1 2 35 π₯ β 3π₯ +
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INECUACIONES DE SEGUNDO GRADO 1) Al resolver el sistema: 1 2 35 π₯ β 3π₯ + >0 2 8 (2x-7)(2x-5) X=1/2 X=5/2 1 2 35 π₯ β 3π₯ + π₯ 2 β 4π₯
π¦ = β(π₯ β 2)2 + 4 π¦ = (π₯ β 2) β 4
π¦ = (2; 4)
π¦ = (2; β4)
β(π₯ β 2)2 + 4 < π¦ < (π₯ β 2)2 + 4 β4 < π¦ < 4 => {1,2,3} => {2} ππ π£πππππππππ ππ πππ πππ’πππππ => {1,3} (1, β3)(1, β2)(1, β1)(1,0)(1,1)(1,2)(1,3) (3, β3)(3, β2)(3, β1)(3,0)(3,1)(3,2)(3,3) Rpta: 14 pares ordenadas 9) Resuelve la inecuaciΓ³n |π₯ β 2|2 β 2|x β 2| β 15 > 0 π2 β 2π β 15 > 0 π
+3
π
β5
(π + 3)(π β 5) > 0 π=3
π=5
π₯|π₯ β 2| = β3
π₯|π₯ β 2| = 5
π₯=5
π₯ = β3
10) Si el sistema β3
(π β 3)2 β 4(4)(4) (π β 4)(π + 5) > 0 β7 < π < 11 12) Al resolver el sistema: π₯ 2 + 8π₯ + 15 < 0 π₯ 2 β 2π₯ β 24 < 0 El conjunto soluciΓ³n es . Hallar el valor de β2b-aβ. π₯ 2 + 8π₯ + 15 < 0 (x + 5)(x β 3)
π₯
5
π₯1 = π₯ β 5
π₯
3
π₯2 = π₯ + 3
π₯ 2 β 2π₯ β 24 < 0
(π₯ β 6)(π₯ + 4)
π₯
β6
π₯1 = 6
π₯
+4
π₯2 = β4
β2b-aβ 2(-3)-(-4) -6+4 -2
14)ΒΏCuΓ‘ntos pares ordenados (a, b)de coordenadas enteras satisfacen el sistema? π¦ β€π₯ 2
π¦ β€ 2π₯
π¦ + 1 > π₯2
π¦ > π₯2 β 1
π₯ 2 β 1 < π¦ < 2π₯ β2π₯. π₯ 2 β 1 < 0 π₯ 2 β 2π₯ β 1 < 0 βπ 2 β 4π 2 + β4 β 4(1) β 1 2π 2 π₯1 =
2 β β8 π₯2 = β0,41 2 π₯ = 2,41
16) La soluciΓ³n de la inecuaciΓ³n: βπ₯ 2 + 8π₯ β 7 > 0 π₯ 2 β 8π₯ + 7 > 0
(π₯ β 7)(π₯ β 1)
π₯
β7
π₯1 = 7
π₯
β1
π₯2 = 1
C.S.= 1 0 X
+7 X
- 2
(x=-7)(x=2) π₯ 2 β π₯ β 12 < 0 X
-4 X
3
(x=+4)(x=-3) Cs
21) Resolver el sistema: 5X-12π₯ 2 + 3 Y