Question

Rozwiąż równania

1.

a) (x-1)³+(2x+3)³=27x³+8

b) (x²-5x+7)²-(x-2)(x-3)=1

c) (x+1)(x²+2)+(x+2)(x²+1)=2

d) (x²+2x)²-(x+1)²=55

2.

x⁷-5x⁵+4x³=0

Answer 1

1 a)

(x - 1)³+(2x + 3)³ = 27x³ + 8

(x - 1)³+(2x + 3)³ = (3x)³+ 2³

(x-1+2x+3)[(x-1 )² - (x-1)(2x+3) + (2x+3)²] = (3x+2)[(3x)²- 3x·2 + 2²]

(3x+2)[x²-2x+1 - (2x²+3x-2x-3) + 4x²+12x+9] = (3x+2)(9x²-6x+4)

(3x+2)[x²-2x+1 - (2x²+x-3) + 4x²+12x+9] = (3x+2)(9x²-6x+4)

(3x+2)(x²-2x+1-2x²-x+3+4x²+12x+9) = (3x+2)(9x²-6x+4)

(3x+2)(3x²+9x+13) = (3x+2)(9x²-6x+4) /:(3x+2)

3x²+9x+13 = 9x²-6x+4

3x²+9x+13-9x²+6x-4=0

-6x²+15x+9=0 /:(-3)

2x²-5x-3=0

Δ = 25+24=49; √Δ=7

Odp. x₁ = -½ i x₂ = 3

 

II sposób

(x - 1)³+(2x + 3)³ = 27x³ + 8

(x - 1)³+(2x + 3)³ = (3x)³+ 2³ lub (x - 1)³+(2x + 3)³= 2³ + (3x)³

Zatem

(x - 1)³+(2x + 3)³=(3x)³+ 2³

x - 1 = 3x i 2x + 3 = 2

x - 3x = 1 i 2x = 2 - 3

-2x - 1 /:(-2) i 2x = - 1 /:2

x = -½ i x = -½ czyli x = -½

(x - 1)³+(2x + 3)³= 2³ + (3x)³

x - 1 = 2 i 2x + 3 = 3x

x = 2+1 i 2x-3x = -3

x = 3 i -x = -3 /·(-1)

x = 3 i x = 3, czyli  x = 3

Odp. x₁ = -½ i x₂ = 3

 

1 b)

(x²-5x+7)² - (x-2)(x-3) = 1

(x²-5x+7)² = 1 + (x-2)(x-3)

(x²-5x+7)² = 1+x²-3x-2x+6

(x²-5x+7)² = x²-5x+7  /:(x²-5x+7)

x²-5x+7 = 1

x²-5x+7-1 = 0

x²-5x+6 = 0

Δ = 25 - 24 = 1; √Δ = 1

Odp. x₁ = 2 i x₂ = 3

 

II sposób

(x²-5x+7)² - (x-2)(x-3) = 1

(x²-5x+7)² - 1 - (x-2)(x-3) = 0

(x²-5x+7)² - 1² - (x-2)(x-3) = 0

Możemy skorzystać ze wzoru a² - b² = (a - b)(a+b)

(x²-5x+7 - 1)(x²-5x+7) - (x-2)(x-3) = 0

(x²-5x+6)(x²-5x+7) - (x-2)(x-3) = 0

Teraz x²+5x+6 rozkładamy na czynniki ze wzoru na postać iloczynową: a²+bx+c = a(x-x₁)(x-x₂)

x²-5x+6 = 0

Δ = 25 - 24 = 1; √Δ = 1

x²-5x+6 = (x-2)(x-3)

Zatem

(x-2)(x-3)(x²-5x+7) - (x-2)(x-3) = 0

(x-2)(x-3)[(x²-5x+7 - 1] = 0

(x-2)(-3)(x²-5x+6) = 0

(x-2)(x-3)(x-2)(x-3) = 0

(x-2)²(x-3)² = 0

(x-2)² = 0 v (x-3)² = 0

x - 2 = 0 v x - 3 = 0

x = 2 v x = 3

Odp. x₁ = 2 i x₂ = 3

 

1 c)

(x+1)(x²+2)+(x+2)(x²+1) = 2

x³+2x+x²+2+x³+x+2x²+2-2 = 0

2x³+3x²+3x+2=0

2x³+2x²+x²+x+2x+2=0

2x²(x+1) + x(x+1) + 2(x+1) = 0

(x+1)(2x²+x+2) = 0

x+1 = 0 v 2x²+x+2=0

x+1 = 0

x = -1

2x²+x+2=0

Δ = 1-16 = -15 < 0 ⇒ równanie nie ma rozwiązań

Odp. x = -1

 

1 d)

(x²+2x)²-(x+1)² = 55

x⁴+4x³+4x²-x²-2x-1-55 = 0

x⁴+4x³+3x²-2x-56 = 0

x⁴+4x³+3x²+12x-14x-56 = 0

x³(x+4)+3x(x+4)-14(x+4) = 0

(x+4)(x³+3x-14) = 0

(x+4)(x³-2x²+2x²-4x+7x-14) = 0

(x+4)[x²(x-2)+2x(x-2)+7(x-2)] = 0

(x+4)[(x-2)(x²+2x+7)] = 0

(x+4)(x-2)(x²+2x+7) = 0

x+4 = 0 v x-2 = 0 v x²+2x+7 = 0

x+4 = 0

x = -4

x-2 = 0

x = 2

x²+2x+7 = 0

Δ = 4 - 28 = - 24 < 0 ⇒ równanie nie ma rozwiązań

Odp. x₁ = -4 i x₂ = 2

 

II sposób

(x²+2x)²-(x+1)² = 55

(x²+2x)²-(x+1)² = 64 - 9

(x²+2x)²-(x+1)² = 8² - 3³

(x²+2x)²-(x+1)² = 8² - 3³

x²+ 2x = 8 i x+1 = 3 lub x+ 1 = - 3

x²+ 2x - 8 = 0

Δ = 4 + 32 = 36; √Δ = 6

x+1 = 3

x = 3-1

x = 2

x + 1 = - 3

x = -3-1

x = -4

Odp. x₁ = -4 i x₂ = 2

 

Zad. 2

x⁷-5x⁵+4x³=0

x³(x⁴ - 5x² + 4) = 0

x³ = 0 v x⁴ - 5x² + 4 = 0

x³ = 0

x = 0

x⁴ - 5x² + 4 = 0

x² = t

t² - 5t  + 4 = 0

Δ = 25 - 16 = 9; √Δ = 3

t₁ = 1

x² = 1

x = 1 v x = -1

t₂ = 4

x² = 4

x = 2 v x = -2

 

Odp. x₁ = 0, x₂ = 1, x₃ = -1, x₄ = 2, x₅ = -2