Question

KRÓTKIE ZADANIE W ZAŁĄCZNIKI, PROSZĘ O POMOC

Answer 1

1)

x³+2x²-9x-18=0

x²(x+2)-9(x+2)=0

(x+2)(x²-9)=0

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

x=-2 ∨ x=3 ∨ x=-3

 

2)

x³+2x²-4x-8≥ 0

x²(x+2)-4(x+2) ≥0

(x+2)(x²-4) ≥ 0

(x+2)(x-2)(x+2) ≥ 0

x=-2 ∨ x=2

x∈ <-2;2> ∨ <2;∞)

 

3)

a)

x∈R\{½}

 

b)

x∈R\{-1;⅕}

Δ=(-3)²-4*6*(-3)=81

√Δ=9

x₁=(3+9)/12=12/12=1

x₂=(3-9)/12=-6/12=½

 

4)

a)

b)

Answer 2

1.

x^3 + 2x^2 - 9x -18 = 0

x^2(x+2) -9(x+2) = 0

(x+2)(x2 -9) = 0

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

x = -3   v   x = -2   v   x = 3

2.

x^3 + 2x^2 - 4x - 8 >= 0

x^2(x+2) -4(x+2) >= 0

(x+2)(x^2 -4) <= 0

x+2)(x+2(x-2) <= 0

x = -2  v  x = 2

xe <-2;2>  v  <2;oo)

3.

a)

(5x+4)/(2x-1 = 3

2x =/=1

x =/=1/2

xe R\{1/2}

5x+4 = 3(2x-1)

5x+4 = 6x -3

-x = -7

x = 7

b

(2x+1)/(x+1 = (4x+2)/(5x-1)

xe R\{-1; 1/5} 

(2x+1)(5x-1) = (x1)(4x+2)

10x^2 - 2x + 5x -1 = 4x^2 + 2x + 4x + 2

6x2 -3x -3 = 0

D(delta) = b2 -4ac = 9=72 = 81

VD = 9

x1 =(3-9)/12 = -1/2 = -0,5

x2 =(3+9)/12 = 1

4.

a) 

1/(2x-3) + 4/(x-4) =[(x-4)+4(2x-3)]/[(2x-3)(x-4)] =(x-4+8x-12)/(2x^2 -8x-3x+12) =

= (9x-16)/2x^2 -11x+12) 

b)

3/x +4/x-4) = [3(x-4)+4x]/[x(x-4)] = (3x-12+4x)/(x^2 -4x) = (7x-12)/(x^2 -4x)