Algebra 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 v 3.0 Equazioni di primo grado numeriche frazionarie 3 =0 π₯π₯ − 2 5 4− =0 π₯π₯ 6 −2=4 π₯π₯ 7π₯π₯ + 14 =0 π₯π₯ − 2 1 1 + =4 π₯π₯ 2 3 +1=0 π₯π₯ − 10 3π₯π₯ − 16 5 = π₯π₯ 3 4 1 = 3π₯π₯ − 1 2 + π₯π₯ 1 42 4 οΏ½9 − οΏ½ = − 6 3 π₯π₯ π₯π₯ ππππππππππππππππππππππ π₯π₯ = π₯π₯ = 1 π₯π₯ = −2 π₯π₯ = 2 7 π₯π₯ = 7 π₯π₯ = 12 π₯π₯ = −9 π₯π₯ = 2 2π¦π¦ − 3π¦π¦ 2 5 = − 3π¦π¦ π¦π¦ + 1 π¦π¦ + 1 π¦π¦ = 1 4 π₯π₯ 4 π₯π₯ π₯π₯ 2 − 2π₯π₯ οΏ½ + οΏ½:οΏ½ − οΏ½ = π₯π₯ 4 π₯π₯ 4 16 − π₯π₯ 2 2 2 3 3 οΏ½1 − οΏ½ + οΏ½1 − οΏ½ = 1 3 ππ 2 ππ 2 1 = π₯π₯ + 1 π₯π₯ − 3 2 3 = π₯π₯ + 1 π₯π₯ − 1 π¦π¦ − 2 π¦π¦ − 4 = π¦π¦ + 1 π¦π¦ + 3 π₯π₯ + 1 π₯π₯ − 1 + =2 π₯π₯ − 2 π₯π₯ + 2 4(ππ − 3) 3 −4= ππ + 3 ππ − 3 1 1 + =1 1 − π₯π₯ π₯π₯ − 1 1 2 =− π₯π₯ − 3 π₯π₯ + 5 π₯π₯ − 1 π₯π₯ + 1 8 − + 2 =0 π₯π₯ + 1 π₯π₯ − 1 π₯π₯ − 1 4 4 = π₯π₯ − 4 π₯π₯ + 4 5 4 π₯π₯ = −8 ππ = 5 π₯π₯ = 7 π₯π₯ = −5 π¦π¦ = 1 2 ππππππππππππππππππππππ ππ = 7 3 π₯π₯ = 1 3 ππππππππππππππππππππππ π₯π₯ = 2 © 2016 - www.matematika.it ππππππππππππππππππππππ 1 di 5 Equazioni di primo grado numeriche frazionarie Algebra 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 v 3.0 1 1 4 + = π₯π₯ 2 − 3π₯π₯ π₯π₯ 2 + π₯π₯ π₯π₯ 2 − 2π₯π₯ − 3 π¦π¦ + 2 π¦π¦ + 1 = π¦π¦ − 4 π¦π¦ − 3 ππππππππππππππππππππππ π¦π¦ = 1 π§π§ + 1 2π§π§ − 2 π§π§ + 1 − = π§π§ − 1 π§π§ + 1 1 − π§π§ 1 + 3π₯π₯ 4 6π₯π₯ + = π₯π₯ π₯π₯ − 2π₯π₯ 2 2π₯π₯ − 1 2π¦π¦ 1 − π¦π¦ 1 − −2= 2 π¦π¦ − 2 2π¦π¦ − π¦π¦ π¦π¦ π§π§ = 0 π₯π₯ = −5 π¦π¦ = − π₯π₯ + 1 3 + 6π₯π₯ 3π₯π₯ − 5π₯π₯ 2 + 6 − = (π₯π₯ + 1)(π₯π₯ − 1) π₯π₯ − 1 π₯π₯ + 1 1 6 5 − 2 = 2 − 3π§π§ 3π§π§ − 2π§π§ π§π§ 1 1 2 − = 2 π₯π₯ − 4 (π₯π₯ + 2)(π₯π₯ − 1) (π₯π₯ − 1)(π₯π₯ − 2) 1 + π¦π¦ π¦π¦ + 1 2(π¦π¦ 2 + 3) 1 + = − π¦π¦ + 3 π¦π¦ − 3 π¦π¦ 2 − 9 π¦π¦ + 3 1 π₯π₯ + 2 π₯π₯ 2 − 1 π₯π₯ − 2 π₯π₯ ππππππππππππππππππππππ π§π§ = 1 4 π₯π₯ = − 3 2 ππππππππππππππππππππππ π₯π₯ = 1 =1 π₯π₯ + 3 = π₯π₯ − 2 π₯π₯ + 4 π₯π₯ = π₯π₯ π₯π₯ 2 = − π₯π₯ − 1 2(π₯π₯ + 1) π₯π₯ + 1 5π₯π₯ 3π₯π₯ = 1+ 2π₯π₯ + 3 2π₯π₯ − 3 2π₯π₯ − 4 π₯π₯ 1 = − 2π₯π₯ + 2 π₯π₯ + 1 π₯π₯ 1 4 1 + 3π₯π₯ 3 οΏ½ (π₯π₯ − 1) + οΏ½ = 2 π₯π₯ 2 2 1 1 − 5ππ + = 1 − 2ππ 2ππ − 1 1 − 4ππ2 2 6 32 − = π§π§ + 2 3 − π§π§ π§π§ + 6 − π§π§ 2 1 1 1 − = π₯π₯ 2 + 4π₯π₯ + 3 π₯π₯ 2 − 2π₯π₯ − 3 π₯π₯ 2 − 9 1 1 2 − = π₯π₯ 2 − 1 π₯π₯ − π₯π₯ 2 π₯π₯ 2 + π₯π₯ 4 3π§π§ 9π§π§ − = π§π§ + 1 π§π§ − 3 9 − 3π§π§ 3 2 11 2 π₯π₯ = − π₯π₯ = 3 8 2 5 π₯π₯ = 1 π₯π₯ = 6 ππ = 0 π§π§ = − 19 4 π₯π₯ = −7 ππππππππππππππππππππππ ππππππππππππππππππππππ © 2016 - www.matematika.it 2 di 5 Equazioni di primo grado numeriche frazionarie Algebra 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 v 3.0 1 2 π₯π₯ + 5 − = 2 2π₯π₯ − 4 π₯π₯ + 2 3π₯π₯ − 12 π₯π₯ = 4 + π₯π₯ 3 + π₯π₯ 5 − = 2 π₯π₯ − 3 π₯π₯ − 2 π₯π₯ − π₯π₯ − 6 ππππππππππππππππππππππ 10 π₯π₯ + =1 2 (π₯π₯ − 5) π₯π₯ − 1 20 11 ππππππππππππππππππππππ π₯π₯ + 5 3π₯π₯ 6 − 2π₯π₯ 2 − = π₯π₯ + 3 π₯π₯ + 2 π₯π₯ 2 + 5π₯π₯ + 6 π₯π₯ = 2 1 3 2 = − π₯π₯ 2 − 9 π₯π₯ + 3 2π₯π₯ + 6 π₯π₯ = 2 − π₯π₯ 1 − 3π₯π₯ + =2 3π₯π₯ + 6 2 + π₯π₯ π₯π₯ = − 1 4 π₯π₯ − 1 + = 3(π₯π₯ + 4) 3π₯π₯ 8π₯π₯ + 2π₯π₯ 2 π₯π₯ 2 π₯π₯ = −5 2 1 4 − 2 = − π₯π₯ π₯π₯ + π₯π₯ (π₯π₯ − 1)(π₯π₯ + 1) ππππππππππππππππππππππ 2 3 1 = − π₯π₯ 2 − 1 π₯π₯ 2 − 4 π₯π₯ 2 + π₯π₯ − 2 π₯π₯ = −7 3π₯π₯ − 12 =0 π₯π₯ 2 − 16 3 3 1 1 1 1 οΏ½ − οΏ½οΏ½ − οΏ½ + = 2π₯π₯ − 2 2π₯π₯ + 2 2π₯π₯ 2 2π₯π₯ + 2 π₯π₯ 1 1 2 + = 2π₯π₯ − π₯π₯ 2 π₯π₯ 2 − 4 π₯π₯ 2 + 2π₯π₯ 1 1 π§π§ + 1 − = 2π§π§ + 4 4 − 2π§π§ (π§π§ + 2)(π§π§ − 2) 1 3 1 = + 2 2 + 4π₯π₯ + 3 18 − 2π₯π₯ π₯π₯ − 2π₯π₯ − 3 π₯π₯ − 2 1 2 + 2 = 2 3 π₯π₯ − π₯π₯ π₯π₯ − 1 π₯π₯ + π₯π₯ 4 4 6(2π₯π₯ + 5) 1 − = − 2 3π₯π₯ − 4 3π₯π₯ + 4 9π₯π₯ − 16 3π₯π₯ − 4 2(π₯π₯ 2 + 2) π₯π₯ + 1 −1= 2 π₯π₯ − 4 π₯π₯ − 2 π π π π π₯π₯ = 4: ππππππ ππππππππππππππππππππππ π₯π₯ = −5 π₯π₯ = 1 ππππππππππππππππππππππ 5 π¦π¦ 1 + = π¦π¦ 3 − 1 π¦π¦ 2 + π¦π¦ + 1 π¦π¦ − 1 π₯π₯ 2 7 2 7 16 © 2016 - www.matematika.it π¦π¦ = 2 ππππππππππππππππππππππ ππππππ π₯π₯ ≠ 0, π₯π₯ ≠ ±1 ππππππππππππππππππ ππππππππππππππππππππππππππ π₯π₯ = 2 3 π π π π π₯π₯ = ±2: ππππππ ππππππππππππππππππππππ 3 di 5 Equazioni di primo grado numeriche frazionarie Algebra 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 v 3.0 3 π₯π₯ − 3 π₯π₯ + 3 π₯π₯ − 3 =οΏ½ − οΏ½:οΏ½ + 1οΏ½ π₯π₯ + 3 π₯π₯ + 3 π₯π₯ − 3 π₯π₯ + 3 π₯π₯ 2 1 1 1 − 2 =− (π₯π₯ − 2)(π₯π₯ + 3) + 2π₯π₯ − 3 π₯π₯ − 3π₯π₯ + 2 2π₯π₯ π₯π₯ + 1 π₯π₯ − 1 π₯π₯ − 1 π₯π₯ + 1 =οΏ½ − οΏ½:οΏ½ + οΏ½ +1 π₯π₯ − 1 π₯π₯ + 1 π₯π₯ + 1 π₯π₯ − 1 π₯π₯ 2 2 + π₯π₯ π₯π₯ + 1 2(π₯π₯ 2 + 2) + = π₯π₯ + 2 π₯π₯ − 2 (π₯π₯ − 2)(π₯π₯ + 2) π₯π₯ 2 2 1 7π₯π₯ − 11 + 2 = 3 − π₯π₯ − 2 3π₯π₯ + 2π₯π₯ − 1 3π₯π₯ − 4π₯π₯ 2 − 5π₯π₯ + 2 4π₯π₯ − π₯π₯ 2 1 1 + = π₯π₯ 4 − 16 4π₯π₯ − 8 4π₯π₯ + 8 1 1 2 π₯π₯ + 3 : π₯π₯ + 3π₯π₯ − π₯π₯ + 2π₯π₯ − 3 = 2 1 1 3π₯π₯ − 9 3 6 π₯π₯ − 3 1 1 π¦π¦ − 2 + π¦π¦ + 2 =2 1 1 − π¦π¦ + 2 π¦π¦ − 2 π₯π₯ = −1 π₯π₯ = 6 ∀π₯π₯ ≠ ±1 ππππππππππππππππππππππ ππππππππππππππππππππππ π₯π₯ = −1 π₯π₯ = − 6 5 π¦π¦ = −4 1 π₯π₯ + 2 π₯π₯ + 4 =2− 1 π₯π₯ − 4 π₯π₯ − 2 π₯π₯ = 8 9 1 − 2π§π§ 4π§π§ − 6 12 + + =0 π§π§ 2 + 3π§π§ 2π§π§ 2 − 6π§π§ 9 − π§π§ 2 π§π§ = −6 4 4 5 5 − − − = π₯π₯ 1 − π₯π₯ π₯π₯ + π₯π₯ 2 π₯π₯ 2 − 1 π₯π₯ = ππ + 2 8 2ππ ππ − 2 − = − ππ2 − 2ππ ππ2 − 4 ππ2 − 4 ππ2 − 2ππ π₯π₯ 2 2π₯π₯ 12 2π₯π₯ + = 2 2 − 3π₯π₯ 9 − π₯π₯ π₯π₯ + 3π₯π₯ 1 1 1 π§π§ 2 + 5 + + = 5π§π§ + 5 5π§π§ − 5 5 5 + 5π§π§ 2 − 10π§π§ 3 3 3 π₯π₯ − 3π₯π₯ 2 + + = π₯π₯ 2 − π₯π₯ π₯π₯ 2 + π₯π₯ π₯π₯ π₯π₯ − π₯π₯ 3 π₯π₯ − 3 2π₯π₯ − 1 = −5 π₯π₯ + 1 π₯π₯ + 1 3 2 = π₯π₯ − 1 3 ππππππππππππππππππππππ ππππππ π₯π₯ ≠ −3 ππππππππππππππππππ ππππππππππππππππππππππππππ π₯π₯ = − π₯π₯ = π₯π₯ = π₯π₯ = © 2016 - www.matematika.it 3 2 3 7 1 2 8 11 11 2 4 di 5 Equazioni di primo grado numeriche frazionarie Algebra 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 v 3.0 1 1 1 = − 2π₯π₯ − 1 π₯π₯ 2π₯π₯ + 2 π₯π₯ = 2 π₯π₯ 1 π₯π₯ + 1 = + 2π₯π₯ + 6 2 π₯π₯ + 3 π₯π₯ 2 π₯π₯ = − 2π₯π₯ − 6 1 2 = + − 2π₯π₯ − 15 π₯π₯ + 3 π₯π₯ − 5 π₯π₯ 2 π₯π₯ − 3 2π₯π₯ 3 6π₯π₯ 2 − 9π₯π₯ −9 + = 2 4π₯π₯ − 6 3 − 2π₯π₯ 6 − 4π₯π₯ π₯π₯ π₯π₯ 2 π₯π₯ 2 = + +1− 2 π₯π₯ − 3 π₯π₯ + 3 π₯π₯ 2 − 9 π₯π₯ − 9 π₯π₯ 2 = 2 +1 π₯π₯ − 5 π₯π₯ − 25 π₯π₯ 2π₯π₯ + 1 3 = + 2 π₯π₯ + 1 2π₯π₯ + 2 π₯π₯ − 1 6π₯π₯ + 4 4π₯π₯ 1 2 − = − 4π₯π₯ 2 + 4π₯π₯ + 1 4π₯π₯ 2 − 1 2π₯π₯ + 2 4π₯π₯ 2 − 1 π₯π₯ 2 3π₯π₯ 2π₯π₯ 5π₯π₯ + 6 + + = +1 2 π₯π₯ − 5π₯π₯ − 14 π₯π₯ + 2 π₯π₯ − 7 π₯π₯ + 2 π₯π₯ + 7 π₯π₯ 2 − 1 1 3 − 2π₯π₯ 2 + π₯π₯ = − − π₯π₯ + 2 π₯π₯ + 1 π₯π₯ + 2 π₯π₯ 2 + π₯π₯ + 2 π₯π₯ 2 − 1 π₯π₯ 3π₯π₯ 2 = − π₯π₯ 2 − π₯π₯ π₯π₯ − 4 π₯π₯(π₯π₯ − 1)(π₯π₯ − 4) 1 3 2 = − − −π₯π₯ 2 + 2π₯π₯ − 1 π₯π₯ − 1 3π₯π₯ − 3 ππππππππππππππππππππππ π₯π₯ = − π₯π₯ = − 2π₯π₯ π₯π₯ 2 3π₯π₯ − π₯π₯ 2 π₯π₯ − π₯π₯ 2 − + = 2 π₯π₯ − 3 π₯π₯ − 2 2 − π₯π₯ π₯π₯ − 5π₯π₯ + 6 2π₯π₯ + 1 5 2 + = 2 π₯π₯ + 1 1 − π₯π₯ π₯π₯ − 1 4 − 2π₯π₯ 5 3π₯π₯ + 2 + = π₯π₯ + 2 1 − π₯π₯ π₯π₯ − 1 5 1 3 20π₯π₯ 2 + 27 1 − − = − 2 2 2 2 2 4 2 (4π₯π₯ − 1) 4π₯π₯ − 4π₯π₯ + 1 (2π₯π₯ + 1) 16π₯π₯ − 8π₯π₯ + 1 4π₯π₯ − 1 © 2016 - www.matematika.it 7 6 23 5 π₯π₯ = −5 π₯π₯ = − π₯π₯ = − 3 4 56 17 ππππππππππππππππππππππ ππππππππππππππππππππππ 14 11 π₯π₯ = − π₯π₯ 1 2 = + π₯π₯ + 1 3π₯π₯ + 3 3 5+ π₯π₯ = −7 π₯π₯ = 4π₯π₯ − 1 3π₯π₯ + 5 1 − 3π₯π₯ =− + 2 4π₯π₯ − 1 1 − 2π₯π₯ 2π₯π₯ + 1 5 2 π₯π₯ = 3 5 14 π₯π₯ = 0 π₯π₯ = 4 π₯π₯ = −14 π₯π₯ = 9 8 5 di 5