Algebra 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 v 3.1 Disequazioni biquadratiche π₯π₯ 4 − 2π₯π₯ 2 + 1 ≤ 0 π₯π₯ = ±1 π₯π₯ 4 − π₯π₯ 2 − 2 < 0 −√2 < π₯π₯ < √2 π₯π₯ 4 − 3π₯π₯ 2 − 4 ≥ 0 π₯π₯ ≤ −2 ∨ π₯π₯ ≥ 2 4π₯π₯ 4 − 17π₯π₯ 2 + 18 ≥ 0 π₯π₯ ≤ − 5π₯π₯ 4 − 16π₯π₯ 2 + 3 > 0 π₯π₯ < −√3 ∨ − 10π₯π₯ 4 − 11π₯π₯ 2 + 3 ≥ 0 π₯π₯ ≥ − π₯π₯ 4 − 4π₯π₯ 2 + 3 > 0 π₯π₯ < −√3 ∨ −1 < π₯π₯ < 1 ∨ π₯π₯ > √3 (π₯π₯ 2 + 1)2 − 4(2 − π₯π₯ 2 ) ≤ 0 −1 ≤ π₯π₯ ≤ 1 2π₯π₯ 4 − 3π₯π₯ 2 + 1 ≤0 π₯π₯ 4 − 2π₯π₯ 2 + 1 −1 < π₯π₯ ≤ − 2π₯π₯ 4 − 31π₯π₯ 2 − 16 < 0 3 2 ∨ −√2 ≤ π₯π₯ ≤ √2 ∨ π₯π₯ ≥ 3 2 −4 < π₯π₯ < 4 √5 5 < π₯π₯ < √5 5 10π₯π₯ 4 + 11π₯π₯ 2 + 3 < 0 ππππππππππππππππππππππ π₯π₯ 4 − 5π₯π₯ 2 + 4 < 0 −2 < π₯π₯ < −1 ∨ 1 < π₯π₯ < 2 (π₯π₯ 2 + 1)(π₯π₯ 2 − 4) ≥ 0 π₯π₯ ≤ −2 ∨ π₯π₯ ≥ 2 (2π₯π₯ 2 − 3)(π₯π₯ 2 + 1) + π₯π₯ 2 (1 − π₯π₯ 2 ) < 0 − √3 < π₯π₯ < √3 4π₯π₯ 4 − 13π₯π₯ 2 − 75 ≥0 π₯π₯ 4 − 5π₯π₯ 2 + 4 (π₯π₯ 2 − 2)(1 + π₯π₯ 2 ) (π₯π₯ 2 + 1) < (1 − 2π₯π₯ 2 ) 3(2π₯π₯ 2 − 1) (π₯π₯ 2 − 3)(π₯π₯ 2 − 2) < 0 © 2016 - www.matematika.it √15 √2 √2 √15 ∨− ≤ π₯π₯ ≤ ∨ π₯π₯ ≥ 5 2 2 5 4 π₯π₯ ≤ − − ∨ π₯π₯ > √3 4 √2 2 ∨ √2 2 ≤ π₯π₯ < 1 5 5 ∨ −2 < π₯π₯ < −1 ∨ 1 < π₯π₯ < 2 ∨ π₯π₯ ≥ 2 2 √2 √2 < π₯π₯ < 2 2 −√3 < π₯π₯ < −√2 ∨ √2 < π₯π₯ < √3 1 di 2 Algebra 18 19 20 21 22 23 24 25 26 27 28 29 30 31 v 3.1 Disequazioni biquadratiche (2π₯π₯ 2 + 3)(π₯π₯ 2 − 2) ≤ 0 −√2 ≤ π₯π₯ ≤ √2 π₯π₯ 4 − 9π₯π₯ 2 + 20 ≤ 0 −√5 ≤ π₯π₯ ≤ −2 ∨ 2 ≤ π₯π₯ ≤ √5 π₯π₯ 4 − 12π₯π₯ 2 + 27 ≥ 0 π₯π₯ ≤ −3 ∨ −√3 ≤ π₯π₯ ≤ √3 ∨ π₯π₯ ≥ 3 π₯π₯ 4 − 8π₯π₯ 2 − 9 > 0 π₯π₯ < −3 ∨ π₯π₯ > 3 π₯π₯ 4 + π₯π₯ 2 + 1 < 0 ππππππππππππππππππππππ 5π₯π₯ 4 + 3π₯π₯ 2 + √2 > 0 ∀π₯π₯ ∈ β π₯π₯ 4 + 2π₯π₯ 2 − 3 ≤ 0 −1 ≤ π₯π₯ ≤ 1 (π₯π₯ 2 − 1)(16π₯π₯ 2 − 1) < 0 −1 < π₯π₯ < − π₯π₯ 4 − 14π₯π₯ 2 − 32 < 0 −4 < π₯π₯ < 4 4π₯π₯ 4 − 5π₯π₯ 2 + 1 ≥ 0 1 1 π₯π₯ ≤ −1 ∨ − ≤ π₯π₯ ≤ ∨ π₯π₯ ≥ 1 2 2 (π₯π₯ 2 − 1)2 + 4π₯π₯ 2 (1 − π₯π₯ 2 ) ≤ 0 π₯π₯ ≤ −1 ∨ π₯π₯ ≥ 1 (π₯π₯ 2 + 2)2 − (2 + 7π₯π₯ 2 ) > 0 1 1 ∨ < π₯π₯ < 1 4 4 π₯π₯ < −√2 ∨ −1 < π₯π₯ < 1 ∨ π₯π₯ > √2 π₯π₯ 4 + π₯π₯ 2 − 2 >0 π₯π₯ 4 − 3π₯π₯ 2 − 4 π₯π₯ < −2 ∨ −1 < π₯π₯ < 1 ∨ π₯π₯ > 2 π₯π₯ 4 + 2ππ2 π₯π₯ 2 − 3ππ4 > 0 π₯π₯ < −|ππ| ∨ π₯π₯ > |ππ| © 2016 - www.matematika.it 2 di 2