Question

Se considera expresia F(x)= $\frac{x-6}{x+2}$($\frac{ x^{2} -5x+6}{ x^{2} -9x+18}$-$\frac{ x^{2} +9x+20}{ x^{2} +11x+30}$) : $\frac{6x+18}{ x^{2} +7x+6}$

a)Aratati ca F(x)=$\frac{x+1}{x+3}$
b)Rezolvati ecuatia F(x)+4=$\frac{x+5}{x+3}$
 E URGENT! 

Answer 1

a)
x^2-5x+6=(x-2)(x-3)
x^2-9x+18=(x-3)(x-6)

(x^2-5x+6)/(x^2-9x+18)=(x-2)/(x-6)

x^2+9x+20=(x+4)(x+5)
x^2+11x+30=(x+5)(x+6)

(x^2+9x+20)/(x^2+11x+30)=(x+4)/(x+6)

(.......paranteza din exercitiu devine......)=(x-2)/(x-6) -(x+4)/(x+6)=
=[(x-2)(x+6)-(x+4)(x-6)]/[(x-6)(x+6)]=
=(x^2+4x-12-x^2+2x+24)/[(x-6)(x+6)]=(6x+12)/[(x-6)(x+6)]=6(x+2)/[(x-6)(x+6)]

(6x+18)/(x^2+7x+6)=[6(x+3)]/[(x+1)(x+6)]

Conditii de existentialitate: x diferit de {-7,-6,-5,-3,-2,-1,3,6}

F(x)=[(x-6)/(x+2)]*6(x+2)/[(x-6)(x+6)]:[6(x+3)]/[(x+1)(x+6)]=

={[6(x-6)(x+2)]/[(x+2)(x-6)(x+6)]}*{[(x+1)(x+6)]/[6(x+3)]}=

=[6/(x+6)]*{[(x+1)(x+6)]/[6(x+3)]}=[6(x+1)(x+6)]/[6(x+6)(x+3)]=

=(x+1)/(x+3)

b)
(x+1)/(x+3)+4=(x+5)/(x+3),

x diferit de -3

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

x=-2