Question

If x^5 = 1 , find the value of the fractional expression \dfrac{x}{1+x^2}+\dfrac{x^2}{1+x^4}+\dfrac{x^3}{1+x}+\dfrac{x^4}{1+x^3} .

Collected in the board: Fractions

Steven Zheng posted 2 months ago


Answer

\because x^5=1

\therefore x^6 = x , x^7 = x^2

\dfrac{x}{1+x^2}+\dfrac{x^2}{1+x^4}+\dfrac{x^3}{1+x}+\dfrac{x^4}{1+x^3}

= \dfrac{x}{1+x^2}\cdotp \dfrac{x^4}{x^4} +\dfrac{x^2}{1+x^4}\cdotp \dfrac{x^3}{x^3} +\dfrac{x^3}{1+x}\cdotp \dfrac{x^2}{x^2} +\dfrac{x^4}{1+x^3}\cdotp \dfrac{x}{x}

=\dfrac{x^5}{x^4+x^6} +\dfrac{x^5}{x^3+x^7}+\dfrac{x^5}{x^2+x^3}+\dfrac{x^5}{x+x^4}

=\dfrac{1}{x^4+x}+\dfrac{1}{x^3+x^2}+\dfrac{1}{x^2+x^3}+\dfrac{1}{x+x^4}

=\dfrac{2}{x}\cdotp\Big( \dfrac{1}{1+x^3}+\dfrac{1}{x+x^2}\Big)

=\dfrac{2}{x}\cdotp\dfrac{x^3+x^2+x+1}{x+x^2+x^4+x^5}

=\dfrac{2}{x}\cdotp\dfrac{x^3+x^2+x+1}{x^4+x^2+x+1}

=2\cdotp \dfrac{x^3+x^2+x+1}{x^3+x^2+x+1}

=2

Steven Zheng posted 2 months ago

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