Question

Rozwiąż równanie:
[tex]x+x^3+x^5+...=\frac{2}3}[/tex]

Answer 1

[tex]x+x^{3}+x^{5}+...=\frac{2}{3} \\a_{1}=x\ oraz\ q=x^{2}\\\frac{a_{1}}{1-q}=\frac{2}{3} \\\frac{x}{1-x^{2}}=\frac{2}{3} \\2(1-x^{2})=3x\\2-2x^{2}=3x\\2x^{3}+3x-2=0\\\\Zal.\\|q| < 1\\|x^{2}| < 1\\x^{2} < 1\\x^{2}-1 < 0\\(x-1)(x+1) < 0\\[/tex]

x ∈ (-1;1)

Δ 9 + 16 = 25

√Δ = √25 = 5

[tex]x_{1} = \frac{-3+5}{4}=\frac{2}{4}=\frac{1}{2} \\x_{1} = \frac{-3-5}{4} =\frac{-8}{4}=-2[/tex]

Wliczając dziedzinę, otrzymujemy:

[tex]x = \frac{1}{2}[/tex]