Question

prosze mi obliczyc dziedzine funkcji bede bardzo wdzieczny :

1)x²-25/x²-8x+15

2)1-x²/x²+2x-3

3)x³+2x²-3x-6/x³+2x²-5x-10

4)x³+x²+x+1/x³+x²-2x-2

Answer 1

1)x²-8x+15≠0

Δ=(-8)²-4*1*15=64-60=4

√Δ=2

x₁=(8+4)/2=6

x₂=(8-4)/2=2

 

x∈R\{2;6}

 

2)x²+2x-3≠0

Δ=2²-4*1*(-3)=4+12=16

√Δ=4

x₁=(-2+4)/2=1

x₂=(-2-4)/2=-3

 

x∈R\{-3;1}

 

3)x³+2x²-5x-10≠0

x²(x+2)-5(x+2)≠0

(x²-5)(x+2)≠0

 

x²-5≠0

x²≠5

x≠√5 ∨ x≠-√5

 

x+2≠0

x≠-2

 

x∈R\{-2;-√5;√5}

 

4)x³+x²-2x-2≠0

x²(x+1)-2(x+1)≠0

(x²-2)(x+1)≠0

 

x²-2≠0

x²≠2

x≠√2 ∨ x≠-√2

 

x+1≠0

x≠-1

x∈R\{-√2;-1;√2}