Question

VA ROG URGENT DAU COROANA!
va rog frumos,imi puteti da imi rezolvati si mie acest exercitiu?am mare nevoie de rezolvare!dar urgent va rog!!!​

Answer 1

Hei! :)

• E(x)=(x²+x+2)(x²+x-4)+8;
• F(x)=(x²-x-2)(x²-x-4)-8

$a)E(x)=(x^{2} +x+2)(x^{2} +x-4)+8\\Notam \ x^{2} +x=t\\=> E(t)=(t +2)(t-4)+8=t^{2} -4t+2t-8+8=t(t-2)\\\\=> E(x)=(x^{2} +x)(x^{2} +x-2)=x(x+1)(x^{2} +2x-x-2)\\=x(x+1)[(x(x+2)-(x+2)]=x(x+1)(x+2)(x-1)\\=> E(x)=(x+1)*x*(x+1)*(x+2)\\\\F(x)=(x^{2} -x-2)(x^{2} -x-4)-8\\Notam \ x^{2} -x=t\\=> F(t)=(t-2)(t-4)-8=t^{2} -4t-2t+8-8=t(t-6)\\\\=> F(x)=(x^{2} -x)(x^{2} -x-6)=x(x-1)(x^{2} -3x+2x-6)=\\=x(x-1)[(x(x-3)+2(x-3)]=x(x-1)(x-3)(x+2)\\\\\\b) \frac{E(x)}{F(x)}=\frac{(x+1)x(x+1)(x+2)}{x(x-1)(x-3)(x+2)}$$=\frac{x+1}{x-3}$

$c) \frac{x+1}{x-3}\ apartine\ lui\ Z => x+1 | x-3\ ;\ x-3|x-3\\=> x-3| x+1-x+3 => x-3|4\\=> x-3=(-1; -2; -4; 1; 2; 4)\\=> x=(-1; 1; 2; 4; 5; 7)$

Answer 2

Răspuns:

Explicație pas cu pas:

Sper că nu am greșit pe undeva... Verifică te rog... :))

Succese!