`(3x^(-2)+(2x-1)^(-1))/(6/(x^-1+2)+3x^-1)` Simplify the complex fraction.

Expert Answers
lemjay eNotes educator| Certified Educator


First, apply the negative exponent rule `a^(-m)=1/a^m` .

`=(3/x^2+ 1/(2x-1))/(6/(1/x+2) + 3/x)`

Then, simplify the fraction `6/(1/x+2)` . To do so, multiply it by `x/x` .

`=(3/x^2+ 1/(2x-1))/(6/(1/x+2) *x/x+ 3/x)`

`=(3/x^2+ 1/(2x-1))/((6x)/(1+2x) + 3/x)`

`=(3/x^2+ 1/(2x-1))/((6x)/(2x+1) + 3/x)`

Then, multiply the top and bottom of the complex fraction by the LCD of the four fractions present.  The LCD is `x^2(2x - 1)(2x + 1)` .

`=(3/x^2+ 1/(2x-1))/((6x)/(2x+1) + 3/x)*(x^2(2x-1)(2x+1))/(x^2(2x-1)(2x+1))`

`=(3(2x-1)(2x+1) + x^2(2x+1))/(6x(x^2)(2x-1)+3x(2x-1)(2x+1))`

To simplify it further, factor out the GCF in the numerator. Also, factor out the GCF of the denominator.

`= ((2x+1)(3(2x-1) + x^2))/(3x(2x-1)(2x^2+2x+1))`


`= ((2x+1)(x^2+6x-3))/(3x(2x-1)(2x^2+2x+1))`

Both numerator and denominator are now in factored form. Since there is no common factor between them, no factors are cancelled.

Therefore, the simplified form is

 `(3x^(-2)+(2x-1)^(-1))/(6/(x^(-1)+2)+3x^(-1))= ((2x+1)(x^2+6x-3))/(3x(2x-1)(2x^2+2x+1))` .

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