What is 6/(3x)+(x+5)/(x^3)+(1-2x^2)/(x^4) ?

The solution to 6/(3x)+(x+5)/(x^3)+(1-2x^2)/(x^4) is (2x^3-x^2+5x+1)/(x^4)

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simplify 6/(3x)+(x+5)/(x^3)+(1-2x^2)/(x^4)

simplify

\frac{6}{3 x} + \frac{x + 5}{x ^{3}} + \frac{1 - 2 x ^{2}}{x ^{4}}

\frac{2 x ^{3} - x ^{2} + 5 x + 1}{x ^{4}}

Solution steps

\frac{6}{3 x} + \frac{x + 5}{x ^{3}} + \frac{1 - 2 x ^{2}}{x ^{4}}

Cancel

\frac{6}{3 x} : \frac{2}{x}

= \frac{2}{x} + \frac{x + 5}{x ^{3}} + \frac{1 - 2 x ^{2}}{x ^{4}}

Simplify

\frac{2}{x} + \frac{x + 5}{x ^{3}} + \frac{1 - 2 x ^{2}}{x ^{4}} : \frac{2 x ^{3} + ( x + 5 ) x + 1 - 2 x ^{2}}{x ^{4}}

= \frac{2 x ^{3} + ( x + 5 ) x + 1 - 2 x ^{2}}{x ^{4}}

Simplify

2 x^{3} + (x + 5) x + 1 - 2 x^{2} : 2 x^{3} - x^{2} + 5 x + 1

= \frac{2 x ^{3} - x ^{2} + 5 x + 1}{x ^{4}}

1 + \frac{i}{1 + \frac{i}{1 + \frac{i}{1 + i}}}

tan (θ) cos (θ)

\frac{6}{8} \cdot \frac{3}{5}

sin^{2} (2 t) + cos^{2} (2 t)

What is 6/(3x)+(x+5)/(x^3)+(1-2x^2)/(x^4) ?
The solution to 6/(3x)+(x+5)/(x^3)+(1-2x^2)/(x^4) is (2x^3-x^2+5x+1)/(x^4)