Question

Dau coroana pentru f) si g) !!

Answer 1

x^2-9=(x+3)(x-3)
F) [x/(x+3)(x-3)-1(x-3)/(x+3)(x-3)-2( x+3)/(x+3)(x-3)]:[(x+3)(x-3)/(x+3)+(5-x^2)/(x+3)]=[(x-1(x-3)-2(x+3) ]/(x+3)(x-3):(x^2-9+5-x^2)/(x+3)= (x-x+3-2x-6)/(x+3)(x-3):(-4)/(x+3)= (-2x-3)/(x+3)(x-3):(-4)/(x+3)=(-2x-3)/(x+3)(x-3) •(x+3)/(-4)=(-2x-3)/(x-3)•1/(-4)=-(2x+3)/-4(x-3)=(2x+3)/4(x-3)
G)[(x+7)(x+1)/x(x+1)-(x+3)x/(x+1)x- (2-x)/(x^2+x)]:2(6x+5)/x(x^2+ 2x+1)=(x^2+x+7x+7-x^2-3x-2+x)/ x(x+1) : 2(6x+5)/x(x+1)^2= (6x+5)/x(x+1) • x(x+1)^2/2(6x+5)= 1/1 • (x+1)/2=(x+1)/2

Answer 2

h)
[x/(x²-9)-1/(x+3)-2/(x-3)] :[x-3+(5-x²)/(x+3)]=
=(x-x+3-2x-6)/(x-3)(x+3) :(x²-9+5-x²)/(x+3)=
=(-3-2x)/(x-3)(x+3) ×(x+3)/(-4)=
=(-2x-3)/(x-3)/(-4)
=(2x+3)/4(x-3)

g)
[(x+7)/x-(x+3)/(x+1)-(2-x)/(x²+x)] : (12x+10)/(x³+2x²+x)=
=[(x+7)/x-(x+3)/(x+1)-(2-x)/(x²+x)] : (12x+10)/x(x+1)²=
=[(x+7)(x+1)-x(x+3)-(2-x)]/x(x+1) × x(x+1)²/2(6x+5)=
=(x²+7x+x+7-x²-3x-2+x)/1 × (x+1)/2(6x+5)=
=(6x+5)(x+1)/2(6x+5)=
=(x+1)/2