Question

Cum se descompune x²+40?
Pt ecuatia: x+3 supra x-4 minus x-3 supra x+4 = x²+40 supra x²-16

Answer 1

[tex]\frac{x+3}{x-4}-\frac{x-3}{x+4}=\frac{x^2+40}{x^2-16}\\ \frac{(x+3)(x+4)-(x-3)(x-4)}{(x-4)(x+4)}=\frac{x^2+40}{(x-4)(x+4)}\\ (x+3)(x+4)-(x-3)(x-4)=x^2+40\\ x^2+3x+4x+12-x^2+4x+3x-12=x^2+40\\ x^2+7x+12-x^2+7x-12=x^2+40\\ 14x=x^2+40\\ x^2-14x+40=0\\ x^2-10x-4x+40=0\\ x(x-10)-4(x-10)=0\\ (x-10)(x-4)=0\\ x-10=0\Rightarrow x=10\\ x-4=0\Rightarrow x=4\\ S:x\in \{10,4\}[/tex]

Answer 2

Centru de existenta:
x-4=0
x=0+4
x=4
x+4=0
x=0-4
x=-4
x^2-16=0
(x+4)(x-4)=0
x+4=0
x=0-4
x=-4
x-4=0
x=0+4
x=4
Fractia nu este definita pentru x∈{4;-4}
D=R-{4;-4}
[tex]\frac{x+3}{x-4}-\frac{x-3}{x+4}=\frac{(x+3)(x+4)}{(x-4)(x+4)}-\frac{(x-3)(x-4)}{(x-4)(x+4)}=\frac{(x+3)(x+4)-(x-3)(x-4)}{(x-4)(x+4)}=\\\frac{14x}{(x-4)(x+4)}\\ \frac{x^{2}+40}{x^{2}-16}=\frac{x^{2}+40}{(x+4)(x-4)}\\ \frac{14x}{(x-4)(x+4)}-\frac{x^{2}+40}{(x-4)(x+4)}= \frac{14x-(x^{2}+40)}{(x-4)(x+4)}=\frac{-x^{2}+14x-40}{(x-4)(x+4)}=\\ \frac{-1(x^{2}-14x+40)}{(x+4)(x-4)}\\x^{2}-14x+40=x^{2}-10x-4x+40=\\ x(x-10)-4(x-10)=(x-4)(x-10)\\\frac{-1(x-4)(x-10)}{(x+4)(x-4)}=\frac{-1(x-10)}{x+4}=\frac{10-x}{x+4}=0\\ 10-x=0*(x+4)\\10-x=0\\x=10-0=10[/tex]