Odpowiedź:
zad 1.
cosα = 3/5
cos²α= (3/5)² = 9/25
1 - sin²α = 9/25
sin²α = 1 - 9/25 = 16/25
sinα = √(16/25) = 4/5
tgα = sinα : cosα = 4/5 : 3/5= 4/5 *5/3 = 4/3 = 1 1/3
zad 2
(2x - 1)/3 = 2x/5
5(2x - 1) = 3 * 2x
10x - 5 = 6x
10x - 6x = 5
4x = 5
x = 5/4 = 1 1/4
zad 3
(3x - 2)/x = x ; x ≠ 0
3x - 2 = x * x = x²
x² - 3x + 2 = 0
a = 1 , b = - 3 , c = 2
Δ = b² - 4ac = (-3)² - 4 * 1 * 2 = 9 - 8 = 1
√Δ = √1 = 1
x₁ = ( - b - √Δ)/2a = (3 - 1)/2 = 2/2 = 1
x₂ = ( - b + √Δ)/2a = (3 + 1)/2= 4/2= 2
zad 4
(x³ - 3)/[(x + 1)(3x - 2)
założenie:
x + 1 ≠ 0 ∧ 3x - 2 ≠ 0
x ≠ - 1 ∧ 3x ≠ 2
x ≠ - 1 ∧ x ≠ 2/3
D: x ∈ R \ { - 1 , 2/3 }
