Question

Ma puteti ajuta ?????

Answer 1

Răspuns:

Explicație pas cu pas:

a) (x²+8x+15)/(x²+2x-15)=(x²+3x+5x+15)/(x²+5x-3x-15)=[x(x+3)+5(x+3)]/[x(x+5)-3(x+5)]=(x+3)(x+5)/(x+5)(x-3)=(x+3)/(x-3)

domeniul de definitie: numaratorul trebuie sa fie ≠ 0

⇒  (x+5)(x-3)≠0 ⇒   x∈R\{-5; 3}

b) (x²-4x-12)/(x²-6x-16)=(x²-4-4x-8)/(x²-4-6x-12)=

[(x+2)(x-2)-4(x+2)]/[(x+2)(x-2)-6(x+2)]=(x+2)(x-6)/(x+2)(x-8)=(x-6)/(x-8)

domeniul de definitie: (x+2)(x-8)≠0  ⇒    x∈R\{-2;8}

c) (x²+x-6)/(x²-x-2)=(x²-4+x-2)/(x²-1-x-1)=[(x+2)(x-2)+(x-2)]/[(x+1)(x-1)-(x+1)]=

(x-2)(x+3)/(x+1)(x-2)=(x+3)/(x+1)    

domeniul de definitie: (x+1)(x-2)≠0  ⇒x∈R\{-1;2}

d) (x²-6x+8)/(x²+2x-24)=(x²-2x-4x+8)/(x²-16+2x-8)=

[x(x-2)-4(x-2)]/[(x+4)(x-4)+2(x-4)]=(x-2)(x-4)/(x-4)(x+6)=(x-2)/(x+6)

domeniul de definitie: (x-4)(x+6)≠0  ⇒x∈R\{-6;4}                

e) (x²+2x-8)/(x²-2x-24)=(x²-4+2x-4)/(x²-16-2x-8)=

[(x+2)(x-2)+2(x-2)]/[(x+4)(x-4)-2(x+4)]=(x-2)(x+4)/(x+4)(x-6)=(x-2)/(x-6)

domeniul de definitie: (x+4)(x-6)≠0  ⇒x∈R\{-4;6}

f) (x²+2x-15)/(x²+x-20)=(x²-9+2x-6)/(x²-25+x+5)=

[(x+3)(x-3)+2(x-3)]/[(x+5)(x-5)+(x+5)]=(x-3)(x+5)/(x+5)(x-4)=(x-3)/(x-4)

domeniul de definitie: (x+5)(x-4)≠0  ⇒x∈R\{-5;4}

g) (x²-6x-16)/(x²-4x-12)=(x²-4-6x-12)/(x²-4-4x-8)=

[(x+2)(x-2)-6(x+2)]/[(x+2)(x-2)-4(x+2)]=(x+2)(x-8)/(x+2)(x-6)=(x-8)/(x-6)

domeniul de definitie: (x+2)(x-6)≠0  ⇒x∈R\{-2;6}

h) (x²+3x-18)/(x²+2x-15)=(x²-9+3x-9)/(x²-9+2x-6)=

[(x+3)(x-3)+3(x-3)]/[(x+3)(x-3)+2(x-3)]=(x-3)(x+6)/(x-3)(x+5)=(x+6)/(x+5)

domeniul de definitie: (x-3)(x+5)≠0  ⇒x∈R\{-5;3}

i) (x²+2x-24)/(x²+4x-12)=(x²-16+2x-8)/(x²-4+4x-8)=

[(x+4)(x-4)+2(x-4)]/[(x+2)(x-2)+4(x-2)]=(x-4)(x+6)/(x-2)(x+6)=(x-4)/(x-2)

domeniul de definitie:/(x-2)(x+6)≠0  ⇒ x∈R\{-6;2}