Question

zad.1.67

Answer 1

Zad.1.67

a)

(2x+4)·(4x-1) = 0

2x+4 = 0  /:2

x+2 = 0

x = -2

lub

4x-1 = 0

4x = 1  /:4

x = 1/4

Odp. x = -2   v   x = 1/4

b)

(2x-1)·x² = 0

2x-1 = 0

2x = 1  /:2

x = 1/2

lub

x² = 0  => x = 0

Odp. x = 0   v   x = 1/2

c) 

(1+x)/(x+5) = 0

D:  R \ {-5}

Mianownik nie może być zerem, zatem:

1+x = 0

x = -1

Odp.x = -1

d)

(5x-5)/(x-1) = 0

D: R \{1}

 

5x-5 = 0  /:5

x-1 = 0

x = 1

Odp. x = 1

e) 

(2x-6)(2-x)/(x-2) = 0

D: R \ {2}

2x-6 = 0  /:2

x-3 = 0

x = 3

lub

2-x = 0

-x = -2

x = 2

Odp. x = 2  v   x = 3

f)

(9-x²)/(x-3) = 0

D: R \ {3}

9-x² = (3+x)(3-x) = 0

3+x = 0

x = -3

lub

3-x = 0

-x = -3

x = 3  - wykluczamy, bo 3∉ D

Odp. x = -3 

g)

(x²-4)/(x+2) = 0

D: R \ {-2}

x²-4 = (x+2)(x-2) = 0

x+2 = 0

x = -2  - wykuczamy, bo -2∉D

lub

x-2 = 0

x = 2

Odp. x = 2

h)

(x²+1)/(x+7) = 0

D: R \ {-7}

x²+1 > 0, równanie sprzeczne

i)

x(x-1)(x-2)/(x²-1) = 0

x²-1 = (x+1)(x-1)

D: R \ {-1, 1}

x = 0

lub

x-2 = 0

x = 2

lub

x²-1 = (x+1)(x-1) = 0

x = -1,  wykluczamy, bo -1 ∉ D

x = 1, wykluczamy, bo 1 ∉ D

Odp. x = 0   v   x = 2

j)

(x²-25)(x²-16)/[(x+4)(x-5)] = 0

D: R \ {-4, 5}

x2-25 = (x+5)(x-5) = 0

x = -5   

lub

x = 5  - wykluczamy, bo 5 ∉ D

x²-16 = (x+4)(x-4) = 0

x = -4  - wykluczamy, bo -4 ∉ D

x = 4

Odp. x = -5   v   x = 4