Odpowiedź:
1.
a)
- 3x² + 5 = 0
Jest to równanie kwadratowe postaci ax² + bx + c = 0
a = - 3 , b = 0 , c = 5
Δ = b² - 4ac = 0² - 4 * (- 3) * 5 = 12 * 5 = 60
√Δ = √60 = √(4 * 15) = 2√15
x₁ = (- b - √Δ)/2a = (0 - 2√15)/(- 6) = - 2√15/(- 6) = √15/3
x₂ = (- b + √Δ)/2a = (- 0 + 2√15)/(- 6) = 2√15/(- 6) = - √15/3
b)
3x² - 1 = 2x²
3x² - 2x² - 1 = 0
x² - 1 = 0
Korzystamy ze wzoru a² - b² = (a - b)(a + b)
x² - 1 = (x - 1)(x + 1)
(x - 1)(x + 1) = 0
x - 1 = 0 ∨ x + 1 = 0
x = 1 ∨ x = - 1
c)
4/3x - 2 = 3
4/3x = 3 + 2 = 5 | * 3
4x = 5 * 3 = 15
x = 15/4 = 3 3/4 = 3,75
d)
(3x² - 1)(x + 1) = 3x² - 1 | : (3x² - 1)
x + 1 = 1
x = 1 - 1 = 0
e)
[(3 + 3 * 3)(2+ 2²) + 2]/[3 * (- 4)]² = [(3 + 9)(2 + 4) + 2]/(- 12)² =
= (12 * 6 + 2)/144 = (72 + 2)/144 = 74/144 = 37/72
2.
a)
3x/√5 = 3x/√5 * √5/√5 = 3x√5/√5²= 3x√5/5
b)
4/(√5 - 1)= 4/(√5- 1) * (√5 + 1)/(√5 + 1) = 4(√5 + 1)/[(√5 - 1)(√5 + 1)]=
= 4(√5 + 1)/(√5² - 1²) = 4(√5 + 1)/(5 - 1) = 4(√5 + 1)/4 = √5 + 1
c)
3a/(4 + √10) = 3a/(4 + √10) * (4 - √10)/(4 - √10) =
= 3a(4 - √10)/[(4 + √10)(4 - √10)]\ = 3a(4 - √10)/(4² - √10²) =
= 3a(4 - √10)/(16 - 10) = 3a(4 - √10)/6 = a(4 - √10)/3
3.
a)
y = - 3x² + 2x + 5
a = - 3 , b = 2 , c = 5
Obliczamy x₁ i x₂
Δ = b² - 4ac = 2² - 4 * (- 3) * 5 = 4 + 12 * 5 = 4 + 60 = 64
√Δ = √64 = 8
x₁ = (-b - √Δ)/2a = (- 2 - 8)/(- 6) = - 10/(- 6) = 10/6 = 1 4/6 = 1 2/3
x₂ = ( - b + √Δ)/2a = (- 2 + 8)/(- 6) = 6/(-6)= - 1
Obliczamy współrzędne wierzchołka
p - współrzędna x = - b/2a = - 2/(- 6) = 2/6 = 1/3
q - współrzędna y wierzchołka = - Δ/4a = - 64/(- 12) = 64/12 =5 4/12 =
= 5 1/3
W - współrzędne wierzchołka = ( 1/3 ; 5 1/3 )
b)
f(x)= 5x² -3x - 10
a = 5 , b = - 3 , c = - 10
Δ = b² - 4ac = (-3)² - 4 * 5 * (-10)= 9 + 200 = 209
√Δ = √209
x₁ = (-b - √Δ)/2a = (3 - √209)/10
x₂ = ( - b + √Δ)/2a = (3+√209)/10
p = -b/2a = 3/10
q = - Δ/4a = - 209/20 = 10 9/20
W = ( 3/10 ; 10 9/20 )
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