Resolvendo livro de matemática - Parte 237 скачать в хорошем качестве

Resolvendo livro de matemática - Parte 237

255 A: divisores de 30 A = {1, 2, 3, 5, 6, 10, 15, 30} B: par ou primo B = {2, 3, 5, 6, 10, 30} P(B) = 3/4 256 n(E) = 36 A: soma 7 ou 9 A = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (3, 6), (6, 3), (4, 5), (5, 4)} n(A) = 10 P(A) = 5/18 257 4b, 3v, 5a P = 1/3 258 CCK CKC KCC P = 3/8 259 6 apt n(E) = C6, 3 = 6!/3!3! n(E) = 20 E = {{a11, a12, a21},..., {}} A: 1 apt em cada andar A = {{a11, a21, a31}} n(A) = C2, 1 * C2, 1 * C2, 1 n(A) = 8 P(A) = 2/5 Conjunto dos números complexos i² = -1 Ex1 x² + 4 = 0 x² = -4 x = +-√-4 x = +-√(-1*4) x = +-(√-1 * √4) x = +-(2√i²) x = +- 2i Ex2 x² - 4x + 5 = 0 ∆ = 16 - 20 ∆ = -4 x = (4 +- √(-4))/2 x' = (4 - 2i)/2 = 2 - i x'' = (4 + 2i)/2 = 2 + i S = {2 - i, 2 + i} 1 a) x² + 9 = 0 x² = -9 x = +-√-9 x = +-3i S = {-3i, 3i} b) x² + 49 = 0 x² = -49 x = +-√(-1*49) x = +-7i S = {+-7i} c) x² + 121 = 0 x² = -121 x = +-√(-1*121) x = +-11i S = {+-11i} d) x² - 6x + 13 = 0 ∆ = 36 - 52 ∆ = -16 x = (6 +- √(-1*16))/2 x' = (6 - √(-1*16))/2 = 3 - 2i x'' = (6 + √(-1*16))/2 = 3 + 2i S = {3 - 2i, 3 + 2i} e) x² + 2x + 2 = 0 ∆ = 4 - 8 ∆ = -4 x = (-2 +- √(-1*4))/2 x' = (-2 - √(-1*4))/2 = -1 - i x'' = (-2+ √(-1*4))/2 = -1 + i S = {-1+-i} f) x² - 2x + 4 = 0 ∆ = 4 - 16 ∆ = -12 x = (2 +- √(-1*3*4))/2 x' = (2 - √(-1*3*4))/2 = 1 - i√3 x'' = (2+ √(-1*3*4))/2 = 1 + i√3 S = {1 +- i√3} 2 4x² - 4x + 5 = 0 ∆ = 16 - 80 ∆ = -64 x = (4± √(-1*64))/8 x' = (4 - √(-1*64))/8 = (1- 2i)/2 x'' = (4+ √(-1*64))/8 = (1 + 2i)/2 S = {(1± 2i)/2} 3 x² - x e x - 5 x² - x = x - 5 x² - 2x + 5 = 0 ∆ = 4 - 20 ∆ = -16 x = (2 +- √(-1*16))/2 x' = (2 - √(-1*16))/2 = 1 - 2i x'' = (2+ √(-1*16))/2 = 1 + 2i S = {1 +- 2i} 4 4 = x + y y = 4 - x xy = 29 x(4 - x) = 29 -x² + 4x - 29 =0 x² - 4x + 29 = 0 ∆ = 16 - 116 ∆ = -100 x = (4 +- √(-1*100))/2 x' = (4 - √(-1*100))/2 = 2 - 5i x'' = (4+ √(-1*100))/2 = 2 + 5i S = {2 +- 5i} A forma algébrica z = a + bi Ex1 z = (k-3) + 6i k - 3 = 0 k = 3 Ex2 z = 6 + (m² - 9)i m² - 9 = 0 m² = 9 m = +-3 1 z = (k + 5) - 4i k = -5 2 z = 1 + (m² - 81)i m = +- 9 3 z = (x + 6) - (y² - 16)i a) número real y² - 16 = 0 y = +-4 b) x = -6 y ≠ +-4 4 z = (x - 1) + (2x - 3)i a) x - 1 0 x 1 b) 2x - 3 0 2x 3 x 3/2 5 z = (4m - 5) + (n - 1)i 4m - 5 = 0 m = 5/4 n - 1 = 0 n = 1 Igualdade de números complexos a + bi = c + di a = c e b = d Ex1 (2x + y) + 6i = 5 + (x + 4y)i 2x + y = 5 x + 4y = 6 2x + y = 5 -2x - 8y = -12 -7y = -7 y = 1 x = 2 1 a - bi = 5 + 2i a = 5 b = -2 2 x - 6 + (2 - y)i = 4 + 3i x - 6 = 4 x = 10 2 - y = 3 y = -1