Question

Exercitiul 16 il vreau rezolvat!

Answer 1

16a. D=R
x³-x²+x+x²-x+1-x³+6x²=x³+3x²+9x-3x²-9x-27-(x³-6x²+12x-8)
6x²+1=6x²-12x-19
12x=-20, x=-20/12=-5/3
b. x-4≠0, x≠4, x+4≠0, x≠-4, x∈R- acolada-4, 4acolada
[tex] \frac{2x-3}{x-4} - \frac{2x+3}{x+4} =0 [/tex]

[tex] \frac{(2x-3)(x+4)-(2x+3)(x-4)}{(x-4)(x+4)} =0 [/tex]

$\frac{2 x^{2} +8x-3x-12-(2 x^{2} -8x+3x-12)}{ x^{2} -16} =0$
[tex] \frac{2 x^{2} +5x-12-2 x^{2} +5x+12}{( x^{2} -16)} =0 [/tex]

[tex] \frac{10x}{ x^{2} -16} =0 10x=0, x=0[/tex]