Riješi 14f(x)=x^-3+2x

$$14 f ( x ) = x ^ { - 3 } + 2 x$$

Izračunaj f

$$f=\frac{1}{7}+\frac{1}{14x^{4}}$$

$$x\neq 0$$

Koraci rješavanja linearne jednadžbe

Jednadžba je u standardnom obliku.

$$14xf=2x+\frac{1}{x^{3}}$$

Podijelite obje strane sa 14x.

$$\frac{14xf}{14x}=\frac{2x+\frac{1}{x^{3}}}{14x}$$

Dijeljenjem s 14x poništava se množenje s 14x.

$$f=\frac{2x+\frac{1}{x^{3}}}{14x}$$

Podijelite 2x+\frac{1}{x^{3}} s 14x.

$$f=\frac{1}{7}+\frac{1}{14x^{4}}$$

Izračunaj x (complex solution)

$$x=\frac{2^{\frac{3}{4}}i\left(7f-1\right)^{-\frac{1}{4}}}{2}$$

$$x=\frac{2^{\frac{3}{4}}\left(7f-1\right)^{-\frac{1}{4}}}{2}$$

$$x=-\frac{2^{\frac{3}{4}}\left(7f-1\right)^{-\frac{1}{4}}}{2}$$

$$x=-\frac{2^{\frac{3}{4}}i\left(7f-1\right)^{-\frac{1}{4}}}{2}\text{, }f\neq \frac{1}{7}$$

Izračunaj x

$$x=\frac{2^{\frac{3}{4}}}{2\sqrt[4]{7f-1}}$$

$$x=-\frac{2^{\frac{3}{4}}}{2\sqrt[4]{7f-1}}\text{, }f>\frac{1}{7}$$