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(2) Equazioni logaritmiche. Logaritmi. 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39. ππππππ1 (π₯π₯ − 2) = 0 2. ππππππ4 (3π₯π₯ − 4) = − π₯π₯ − 2 = −2 3 2. ππππππ3. π₯π₯ = 3. 3 2. π₯π₯ = π₯π₯ =. πππππππ₯π₯ 4 = 1. −1 < π₯π₯ < 0 ∪ π₯π₯ > 0. πππππππ₯π₯−2 9 = 2. π₯π₯ = 5. ππππππ1 −2 = 2 ππππππ3−2π₯π₯. ππππππππππππππππππππππ. 1 = −2 3. ππππππ2 (π₯π₯ + 1) + ππππππ2 3 = ππππππ2 (π₯π₯ − 1) ππππππ3 (3π₯π₯ + 1) − ππππππ3 π₯π₯ = 2. 2 ππππππ2 (π₯π₯ + 2) + ππππππ1 π₯π₯ + 1 = 0 1 ππππππ3 π₯π₯ 2 + 2 = − ππππππ1 2 2 3. 41. v 3.0. π₯π₯ =. 1 6. π₯π₯ = −2. 3. 2 9. 1+√2 π₯π₯ = οΏ½. 2. ππππππ2 (π₯π₯ − 2) − ππππππ4 (3π₯π₯ − 1) = 1. 2. π₯π₯ = 8 + 2√14. 1 2 ππππππ1 οΏ½ π₯π₯ − 1οΏ½ = 2 − ππππππ9 π₯π₯ 4 3 3. ππππππππππππππππππππππ. 3π₯π₯ − 1 =2 π₯π₯. π₯π₯ = −. ππππππ1 ππππππ2 (3π₯π₯ − 1) = 0. π₯π₯ = 1. ππππππ2 ππππ(π₯π₯ + 1) = 1 − ππππππ4 3. equazioni logaritmiche risolubili mediante una posizione. 40. 3 − √3 2. π₯π₯ = ±. 3 ππππππ1 π₯π₯ − 2 = ππππππ2 (π₯π₯ 3 − 1). 3. π₯π₯ =. ππππππππππππππππππππππ. 2. ππππππ2 ππππππ3. 10 3. π₯π₯ = 4. πππππππ₯π₯+1 1 = 0. π₯π₯. 11 8. 2 ππππππ2 π₯π₯ + 5 ππππππ π₯π₯ − 3 = 0. 1 78. 2. π₯π₯ = ππ √3 − 1 π₯π₯ = 10−3 ∪ π₯π₯ = √10. ππππππ2 2 π₯π₯ − 4 ππππππ2 π₯π₯ + 4 = 0. π₯π₯ = 4. © 2016 - www.matematika.it. 2 di 7.
(3) Equazioni logaritmiche. Logaritmi. 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62. v 3.0. 3 ππππππ2 π₯π₯ + ππππππ π₯π₯ = −4. ππππππππππππππππππππππ. 3 ππππππ2 π₯π₯ + 5 ππππππ π₯π₯ = 0. π₯π₯ = 1 ∪ π₯π₯ = 10−3. 2 ππππππ2 π₯π₯ + 3 = 7 ππππππ π₯π₯. π₯π₯ = 103 ∪ π₯π₯ = √10 5. ππππππ32 π₯π₯ + ππππππ3 π₯π₯ − 12 = 0. π₯π₯ = 27 π₯π₯ =. −2ππππ2 π₯π₯ + ππππππ + 1 = 0. π₯π₯ = ππ π₯π₯ =. ππππ3 π₯π₯ − 9ππππππ = 0. ππππππ π₯π₯ (ππππππ π₯π₯ + 1) + 5 ππππππ π₯π₯ = ππππππ2 π₯π₯ + 4 ππππππ π₯π₯ − 7 ππππππ2 2 π₯π₯ + 95 = 8√6 ππππππ2 π₯π₯ 3(ππππππππ + 1) = 5 ππππππ2 π₯π₯. π₯π₯ = ππ. √ππ. 7. π₯π₯ = 10−2. π₯π₯ = 2 4√6−1 ∪ π₯π₯ = 2 4√6+1 3−√69 10. π₯π₯ = 10. 2 =5 ππππππ2 (π₯π₯ + 3). 3+√69 10. ∪ π₯π₯ = 10. π₯π₯ = √2 − 3 ∪ π₯π₯ = 1. 1 2 + =2 ππππππ π₯π₯ ππππππ π₯π₯ + 1. π₯π₯ = 10 ∪ π₯π₯ = √10/10. ππππ π₯π₯ + 1 =1 ππππ2 π₯π₯ + 1. π₯π₯ = 1 ∪ π₯π₯ = ππ. (ππππππ π₯π₯ − 3)(ππππππ π₯π₯ + 3) = 0. π₯π₯ = 103 ∪ π₯π₯ = 10−3. ππππππ5 π₯π₯ − ππππππ25 π₯π₯ = 1. π₯π₯ = 25. 3 3 1 + =− ππππππ4 π₯π₯ 2 + ππππππ4 π₯π₯ ππππππ4 π₯π₯ + 1. π₯π₯ = 2. 4 − 2 ππππππ4 π₯π₯ 5 + 3 ππππππ4 π₯π₯ = ππππππ4 π₯π₯ 3 ππππππ4 π₯π₯ + 5. −2√7±2 √7 4. π₯π₯ = 43. ππππππ22 π₯π₯ − 4 ππππππ2 π₯π₯ + 3 = 0. π₯π₯ = 2 ∪ π₯π₯ = 8. 2ππππ2 (π₯π₯ − 1) − 5 ππππ(π₯π₯ − 1) + 2 = 0. π₯π₯ = √ππ + 1 π₯π₯ = ππ 2 + 1. ππππππ32 (π₯π₯ + 2) − ππππππ3 (π₯π₯ + 2) − 2 = 0. π₯π₯ = −. 3ππππππ12 π₯π₯ − 11ππππππ1 π₯π₯ − 4 = 0 2. 1. π₯π₯ = 1 π₯π₯ = ππ 3 π₯π₯ = ππ −3. ππππ2 π₯π₯ − 2 ππππ π₯π₯ + 1 = 0. 2 ππππππ2 (π₯π₯ + 3) +. 1 81. 3. 5 ∪ π₯π₯ = 7 3. π₯π₯ = √2 π₯π₯ =. 2. © 2016 - www.matematika.it. 1 16. 3 di 7.
(4) Equazioni logaritmiche. Logaritmi. 63 64 65 66 67. ππππ2 (π₯π₯ 2 − 1) + 3ππππ(π₯π₯ 2 − 1) = 0. π₯π₯ = ±οΏ½1 +. ππππππ32 (π₯π₯ − 2) + ππππππ1 (π₯π₯ − 2) = 6. π₯π₯ =. ππππππ22 π₯π₯ − ππππππ2 π₯π₯ 3 + 2 = 0. π₯π₯ = 2 ∪ π₯π₯ = 4. 3. ππππππ22 (2π₯π₯ + 1) − ππππππ1 (2π₯π₯ + 1) = 0 2 πππππππ₯π₯−1 3 − πππππππ₯π₯−1 3 − 2 = 0. π₯π₯ =. ππππππ24 (2π₯π₯ − 3) − 5ππππππ22 (2π₯π₯ − 3) + 4 = 0. 69. 2ππππππ42 π₯π₯(π₯π₯ − 1) + ππππππ1 [π₯π₯(π₯π₯ − 1)]7 + 3 = 0. 71 72 73 74 75. 4. 2 ππππππ2π₯π₯+1 27 + ππππππ2π₯π₯+1 1οΏ½27 − 6 = 0. ππππππ2 (π₯π₯ − 4) − 3 =. 2 ππππππ1 (π₯π₯ − 4) 2. − ππππππ1 (2π₯π₯ + 5) (ππππππ3 (2π₯π₯ + 5) − 1) = 2 3. (ππππππ2 (3π₯π₯ + 4) − 1)2 = ππππππ2 2 2. ππππππ3 2π₯π₯ 2 (ππππππ3 2π₯π₯ 2 − 1) = ππππππ3 2π₯π₯ 2 + 3. 77. ππππππ2 (π₯π₯ − 3)2 (ππππππ2 (π₯π₯ − 3) − 2) − ππππππ1 (π₯π₯ − 3) = 2 2. ππππππ32 (2π₯π₯ 2 − π₯π₯) = 1 equazioni logaritmiche con argomento esponenziale. 79 v 3.0. π₯π₯ = −1 ∪ π₯π₯ = 2 ∪ π₯π₯ =. 1 ± √257 2. π₯π₯ = −. 26 81. √3 − 9 18. π₯π₯ = 6 ∪ π₯π₯ = 8 π₯π₯ = −. 7 ∪ π₯π₯ = 2 3. π₯π₯ = −1 ∪ π₯π₯ = 0 ∪ π₯π₯ = −. 76. 78. 5 7 ∪ π₯π₯ = 2 2 7 13 ∪ π₯π₯ = ∪ π₯π₯ = 4 8 π₯π₯ =. π₯π₯ = −2 ± √5. ππππππ12 [π₯π₯(π₯π₯ + 4)] = ππππππ1 [π₯π₯(π₯π₯ + 4)] 2. 1 4. 4 ∪ π₯π₯ = 1 + √3 3. π₯π₯ = 1 ∪ π₯π₯ =. ππππππ12 (3π₯π₯ + 1) + ππππππ3 (3π₯π₯ + 1)6 = −9 3. 19 ∪ π₯π₯ = 29 9. π₯π₯ = 0 ∪ π₯π₯ = −. 2. 68. 70. 1 ∪ π₯π₯ = ±√2 ππ 3. ππππ 2π₯π₯ = 0. 4 ± 3√2 2. 27 π₯π₯ = ±οΏ½ 2. ∪ π₯π₯ = ±οΏ½. 1 6. π₯π₯ = 7 ∪ π₯π₯ =. 1. √2. 3 2 3 ± √33 ∪ π₯π₯ = 12. π₯π₯ = −1 ∪ π₯π₯ =. +3. π₯π₯ = 0 © 2016 - www.matematika.it. 4 di 7.
(5) Equazioni logaritmiche. Logaritmi. 80 81 82 83 84 85 86 87 88 89. ππππππ5 7π₯π₯ = 1. ππππππ5 2π₯π₯ −1 =. π₯π₯ = ππππππ7 5. 1 4. π₯π₯ = 1 +. ππππ(1 − ππ π₯π₯ ) = 0. ππππππππππππππππππππππ. 1 − ππππππ 3π₯π₯ = ππππππ 2π₯π₯. π₯π₯ =. ππππππ(3π₯π₯ + 1) + ππππππ(3π₯π₯ − 1) = 2. π₯π₯ ππππππ2 3 + ππππππ2 5π₯π₯ = (2π₯π₯ − 1) ππππππ2 5 − π₯π₯ ππππππ2 5 ππππ(22π₯π₯ − 9 β 2π₯π₯ + 21) = 0. ππππππ2 (4π₯π₯ + 2π₯π₯ ) − ππππππ2 2 = 0 ππππ 4π₯π₯. 2 −6. 1 4 ππππππ5 2. ππππ 10 ππππ 6. 1 π₯π₯ = ππππππ3 101 2 π₯π₯ = −. ππππ 5 ππππ 3. π₯π₯ = 2 ∪ π₯π₯ = π₯π₯ = 0. − ππππ 64 = 0. ππππππ2 (2π₯π₯ + 1) + ππππππ2 2 (2π₯π₯ + 1) − 2 = 0. ππππ 5 ππππ 2. π₯π₯ = ±3 π₯π₯ = 0. equazioni logaritmiche di riepilogo 90 91 92 93 94 95 96 97 98 99. v 3.0. ππππππ4 (π₯π₯ 2 + 2) − ππππππ4 (π₯π₯ 2 − 1) = ππππππ4 5 − ππππππ4 (π₯π₯ + 1). ππππππππππππππππππππππ. (ππππππ2 π₯π₯ 2 )2 +9 ππππππ2 π₯π₯ + 2 = 0. 1 1 π₯π₯ = , π₯π₯ = 4 4 √2. π₯π₯ √2 ππππππ(10 − π₯π₯ 2 ) − ππππππ 8 = 2 ππππππ − 2 ππππππ 5 5 ππππππ(π₯π₯ − 1) − 2 β ππππππ(π₯π₯ + 1) − ππππππ 8 = −2 3 2 + =2 ππππππ2 π₯π₯ − 1 ππππππ2 π₯π₯ + 1 ππππππ2 3π₯π₯ − 2 = 0 π₯π₯−1 2. ππππππππππππππππππππππ π₯π₯ = 8, π₯π₯ = π₯π₯ =. 2 ππππππ3 4π₯π₯ = ππππππ1 2−π₯π₯ ππππ 2. π₯π₯ = √2. ππππ 4 ππππ 3. π₯π₯ = 0. 3. π₯π₯ = 1 +. −1=0. 3 1 ππππ 2π₯π₯ +1 − 2 ππππ = 1 2 2. √2 2. 2 ππππ 2. 1 2 π₯π₯ = οΏ½ − 7οΏ½ 3 ππππ 2. −2 πΏπΏπΏπΏπΏπΏ 42−π₯π₯ + 5 πΏπΏπΏπΏπΏπΏ 2π₯π₯+1 = 1. π₯π₯ = ππππππ512 80 © 2016 - www.matematika.it. 5 di 7.
(6) Equazioni logaritmiche. Logaritmi. 100 101 102 103 104 105 106. π₯π₯−1. 2 ππππ 33π₯π₯+1 = ππππ 9. π₯π₯ = −1. π₯π₯ ππππππ2 3 − ππππππ4 9 = 0. π₯π₯ = 1. − ππππ 2π₯π₯ + 2 ππππ 3 = 1 − ππππ 4−π₯π₯ (2π₯π₯ − 1) ππππππ7 2 = 1 + π₯π₯ ππππππ1 4 (π₯π₯ + 1) ππππ 3 − ππππ. 7. 1 1 π₯π₯ = 2 ππππ οΏ½ οΏ½ 9 3. π₯π₯ =. ππππ 9 − 1 ππππ 8. π₯π₯ =. ππππ 14 ππππ 16. π₯π₯ = −1. 1 ππππ 2 + 4 ππππππ2 ππ = 0 π₯π₯. π₯π₯ = −ln2 √2. ππππ 3π₯π₯ + 2 ππππππ3π₯π₯ ππ − 3 = 0. π₯π₯ =. 1 2 ∪ π₯π₯ = ππππ 3 ππππ 3. ππππ 10 ππππ 2 ππππ 10 ∪ π₯π₯ = 3 ππππ 2. π₯π₯ = −. 107. π₯π₯π₯π₯π₯π₯π₯π₯ 2 − 3 ππππππ2π₯π₯ 10 − 2 = 0. 108. 2 − 3π₯π₯ ππππππ4 3π₯π₯ + 2(ππππππ4 9 − ππππππ4 4) = 0. 2 π₯π₯ = ± √3 3. πΏπΏπΏπΏπΏπΏ(24π₯π₯ − 1) + 2 πΏπΏπΏπΏπΏπΏ 3 = (1 − 2π₯π₯) πΏπΏπΏπΏπΏπΏ 4. 4 π₯π₯ = ππππππ16 οΏ½ οΏ½ 3. 109 110 111 112 113 114 115 116 117 118 119 120 v 3.0. οΏ½(π₯π₯ + 1) πΏπΏπΏπΏπΏπΏ 3 − 1οΏ½(1 + πΏπΏπΏπΏπΏπΏ 3π₯π₯ +1 ) = 0 3 − π₯π₯ + ππππππ2 32π₯π₯+1 = 0 3. ππππππ3 οΏ½9π₯π₯+2 − 2οΏ½ = π₯π₯ + 1. π₯π₯ = −1 ±. π₯π₯ = −. 1 πΏπΏπΏπΏπΏπΏ 3. ππππ 24 ππππ 9 − ππππ 2. π₯π₯ = −1. ππππ|3π₯π₯ − 1| = π₯π₯π₯π₯π₯π₯ 9. πΏπΏπΏπΏπΏπΏ(4π₯π₯ − 1) + πΏπΏπΏπΏπΏπΏ 2 = πΏπΏπΏπΏπΏπΏ(22π₯π₯ + 3 β 2π₯π₯+1 − 10) ππππππ 5 − ππππππ(2π₯π₯ − 3) = 1. (2 ππππππ π₯π₯ − 5) ππππππ π₯π₯ = 3 − ππππππ π₯π₯. π₯π₯ = ππππππ3 οΏ½ π₯π₯ = 2 π₯π₯ =. 7 4. π₯π₯ = 10. ππππππ(ππ π₯π₯ + 1) = ππππππ(ππ 2π₯π₯ − 1). ππππππ 5 + (π₯π₯ − 2) ππππππ 4 = ππππππ(4π₯π₯ − 11) 1 3 1 + = 2 2 ππππππ2 π₯π₯ − 2 ππππππ2 π₯π₯ − 1 4. ππππππ π₯π₯ − 2 ππππππ π₯π₯ − 2 ππππππ2 π₯π₯ − 4 + = ππππππ π₯π₯ − 1 ππππππ π₯π₯ − 3 (ππππππ π₯π₯ − 3)(1 − ππππππ π₯π₯). © 2016 - www.matematika.it. 2−√10 2. √5 − 1 οΏ½ 2. ∪ π₯π₯ = 10. 2+√2 2. π₯π₯ = ln 2 π₯π₯ = 2 π₯π₯ =. 1 ∪ π₯π₯ = 32 8. 2. π₯π₯ = 102 ∪ π₯π₯ = 103 6 di 7.
(7) Equazioni logaritmiche. Logaritmi. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140. v 3.0. ππππππ2 (2π₯π₯ − 1) ππππππ2 (2π₯π₯+1 − 2) = 0. π₯π₯ = 1 ∪ π₯π₯ = log 2. ππππππ3 οΏ½π₯π₯ 2 − π₯π₯ = ππππππ3 √2. π₯π₯ = −1 ∪ π₯π₯ = 2. ππππ(ππ 2π₯π₯ − 1) = ππππ(1 − ππ π₯π₯ ). ππππππππππππππππππππππ. ππππππ3 (2π₯π₯ − 1) + ππππππ1 (π₯π₯ − 4) = −1. ππππππππππππππππππππππ. 3. ππππππ1 (π₯π₯ − 1) − ππππππ2 (π₯π₯ + 1) = 3. π₯π₯ =. 2. ππππππ(32π₯π₯ + 2) + ππππππ 1 (3π₯π₯ − 2) = 1 10. 1 1 ππππ(π₯π₯ − 1) + ππππ √3 = [ππππ(5π₯π₯ 2 − 20) − ππππ(π₯π₯ − 2)] 2 2 πππππποΏ½51+√π₯π₯ + 51−√π₯π₯ οΏ½ = 1 πππποΏ½√3π π π π π π π π + πππππππποΏ½ = 0. π₯π₯. =. π₯π₯ π₯π₯ = 1. 3√2 4. π₯π₯ = log 3 5 + √3 ππππππππππππππππππππππ π₯π₯ = 0 ππππππππππππππππππππππ. 2π₯π₯ 2 ππππ π₯π₯ + 5π₯π₯ ππππ π₯π₯ − 3 = 0 ππππππππ. 3 2. ππππππππππππππππππππππ. ππ π₯π₯ = 1 ∪ π₯π₯ = ± + 2ππππ 6. √3 π₯π₯ 2. π₯π₯ = 1. πππππππ₯π₯ π₯π₯ = 0. π₯π₯ = 1. π₯π₯ π₯π₯ = π₯π₯ 3−π₯π₯. ππππππ2 (3 − 2π₯π₯) − 2 ππππππ1 π₯π₯ = 0 ππππππ32 (3π₯π₯ − 1) − ππππππ1 (3π₯π₯ − 1)4 + 3 = 0 3. π₯π₯ − ππππππ4 3π₯π₯ = 1 2 1 1 οΏ½ ππππ 2π₯π₯ − ππππ 3οΏ½ οΏ½ − 4οΏ½ = 0 2 ππππ π₯π₯ ππππππ2. 3 ∪ π₯π₯ = 1 2. π₯π₯ =. 1 ππππ 3. π₯π₯ = 1. 2. π₯π₯π₯π₯π₯π₯π₯π₯ 3 = 1. π₯π₯ =. π₯π₯ =. π₯π₯ = 48 π₯π₯ =. πππππππ₯π₯ 3 = 2 ππππππ1 +1 √3. π₯π₯ =. π₯π₯. © 2016 - www.matematika.it. 4 28 ∪ π₯π₯ = 9 81 9 4 ∪ π₯π₯ = √ππ 2 1 + √5 2. 7 di 7.
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