Question

dada a funcao f(x)= x/x+1 - 1/2x-3
calcule
gostaria da resolucao dessa questao pq dei uma empacada na parte do mmc de expressao algebrica da questao b

Answer 1

a)
f(1)=1/2-(1/-1)=1/2+1=3/2

b)
x/(x+1)-1/(2x-3)=-1/3

mmc =3(x+1)(2x-3)

3x(2x-3)-3(x+1)=-(x+1)(2x-3)
6x²-9x-3x-3=-(2x²-3x+2x-3)
6x²-12x-3=-2x²+3x-2x+3
6x²-12x-3+2x²-3x+2x-3=0
8x²-13x-6=0

a=8
b=-13
c=-6

Δ=b²-4ac
Δ=(-13)²-4(8)(-6)
Δ=169+192
Δ=361

x=(-b±√Δ)/2a=[-(-13)±√361)16=(13±19)/16

x'=(13+19)/16=32/16=2
x"=(13-19)/16=-6/16=-3/8

Answer 2

$f(1)= \frac{1}{1+1} - \frac{1}{2-3} = \frac{1}{2} - \frac{1}{-1} = \frac{1}{2} + \frac{1}{1} = \frac{3}{2}$

b $\frac{x}{x+1} - \frac{1}{2x-3} = - \frac{1}{3} ⇒ \frac{3x}{x-1} - \frac{3}{2x-3} = -1 ⇒<strong></strong> \frac{(2x-3)*3x - 3*(x+1)}{(x+1)*(2x-3)} = -1$
$\frac{6x^{2} - 9x - 3x -3}{2x^{2} - 3x+2x-3} = -1 == \frac{6x^{2} -12x-3}{2x^2-x-3}=-1$
$6x^{2} -12x-3 = -1*(2x^{2} -x-3)$
$8x^{2}-13x-6 = 0$

Resolvendo você encontrará as raízes que serão as repostas, qualquer duvida só falar