Matematyka
piotrek1025
2017-06-24 03:51:39
Rozwiąż równania i nierówności liniowe
Odpowiedź
mamutarower
2017-06-24 10:25:12

1) a) 7x + 4 = 9x - 2 -2x = -6 x = 3 b) 4y - 4 - 6 - 2y + 5 = 0 2y = 5 y = 5/2 c) x/3 - 15/3 = (x - 2)/6  (x - 15)/3 = (x - 2)/6 6(x-15) = 3(x-2) 6x - 90 = 3x - 6 3x = 84 x = 28 d) (3 - 2x)/5 + 40/5 = (5x + 2)/2 - 2x/2 (43 - 2x)/5 = (3x + 2)/2 2(43 - 2x) = 5(3x +2) 86 - 4x = 15x + 10 19x = 76 x = 4 e) 2(x + 2) = 5(4 - 3x) 2x + 4 = 20 - 15x -17x = -16 x = 16/17 f) (3 - 2x)/3 + 3x/3 = -1 (3 + x) = -3 x = -6 g) x + √2 - √2x = 1 x - √2x = 1 - √2 x(1 - √2) = 1 - √2 x = (1 - √2)/( 1 - √2) x = 1 h) 2x - 1 = x√2 + √2 /*√2 2√2x - √2 = 2x + 2 2√2x - 2x = 2 + √2 x(2√2 - 2) = 2 + √2 x = (2+√2)/(2√2-2) * (2√2+2)/(2√2+2) = (4√2 + 4 + 4 + 2√2)/4 = (6√2 + 8)/4= = (3√2 + 4)/2 i) x^2 - 4x + 4 - 3x - 3 > x^2 - 4 - 2 - 7x + 1 > -6 -7x > - 7 x<1  2) a) 7 + 12x + 15 - 15x - 10 ≥ - 5 -3x + 12 ≥ - 5 3x ≤ 17 x ≤ 17/3 c) 12/3 - x/3 > (x - 1)/6 (12 - x)/3 > (x - 1)/6 2(12 - x)/6 - (x - 1)/6 > 0 [2(12 - x) - (x - 1)]/6 > 0 (24 - 2x - x + 1)/6 > 0 -3x + 25 > 0 3x < 25 x < 25/3 d) 2x/6 - (1 - 5x)/6 ≤ 0 (2x - 1 + 5x)/6 ≤ 0 7x - 1 ≤ 0 x ≤ 1/7 e) x^2 - 8x + 16 + x ≤ x^2 - 5 -7x ≤ -5 - 16 x ≥ 3 f) 2(x^2 - 2x + 1) + x^2 - 2x + 1 ≤ x^2 + 2x + 1 2x^2 - 4x + 2 + x^2 - 4x ≤ x^2 2x^2 - 8x + 2 ≤ 0 x^2 - 4x + 1 ≤ 0 ∆ = 16 - 4 = 12 √∆ = √12 = 2√3 x1 = (4 - 2√3)/ 2 = 2 - √3 x2 = (4 + 2√3)/2 = 2 + √3 xe < 2 - √3 ; 2 + √3 > g) 2√5x - 2x < 1 x(2√5 - 2) < 1 x < 1/(2√5 - 2) x < (2√5 + 2)/16 x < (√5 + 2)/8 h) 3x - πx ≤ π - 3 -x(π - 3) ≤ π - 3 -x ≤ 1 x ≥ -1 i) x(3√2 - 1) > 1 + 2√2 x > (1 + 2√2) / (3√2 - 1)  x > (5√2 + 13) / 17

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