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

Saturday, 25 April 2015

$\displaystyle\frac{{{3}{{x}}^{{-{2}}}+{{\left({2}{x}-{1}\right)}}^{{-{1}}}}}{{\frac{{6}}{{{{x}}^{{-{{1}}}}+{2}}}+{3}{{x}}^{{-{{1}}}}}}$ Simplify the complex fraction.

$\displaystyle\frac{{{3}{{x}}^{{-{2}}}+{{\left({2}{x}-{1}\right)}}^{{-{1}}}}}{{\frac{{6}}{{{{x}}^{{-{1}}}+{2}}}+{3}{{x}}^{{-{1}}}}}$

First, apply the negative exponent rule $\displaystyle{{a}}^{{-{m}}}=\frac{{1}}{{{a}}^{{m}}}$ .

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{6}}{{\frac{{1}}{{x}}+{2}}}+\frac{{3}}{{x}}}}$

Then, simplify the fraction $\displaystyle\frac{{6}}{{\frac{{1}}{{x}}+{2}}}$ . To do so, multiply it by $\displaystyle\frac{{x}}{{x}}$ .

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{6}}{{\frac{{1}}{{x}}+{2}}}\cdot\frac{{x}}{{x}}+\frac{{3}}{{x}}}}$

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{{6}{x}}}{{{1}+{2}{x}}}+\frac{{3}}{{x}}}}$

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{{6}{x}}}{{{2}{x}+{1}}}+\frac{{3}}{{x}}}}$

Then, multiply the top and bottom of the complex fraction by the LCD of the four fractions present. The LCD is $\displaystyle{{x}}^{{2}}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}$ .

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{{6}{x}}}{{{2}{x}+{1}}}+\frac{{3}}{{x}}}}\cdot\frac{{{{x}}^{{2}}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}}}{{{{x}}^{{2}}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}}}$

$\displaystyle=\frac{{{3}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}+{{x}}^{{2}}{\left({2}{x}+{1}\right)}}}{{{6}{x}{\left({{x}}^{{2}}\right)}{\left({2}{x}-{1}\right)}+{3}{x}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}}}$

To simplify...

$\displaystyle\frac{{{3}{{x}}^{{-{2}}}+{{\left({2}{x}-{1}\right)}}^{{-{1}}}}}{{\frac{{6}}{{{{x}}^{{-{1}}}+{2}}}+{3}{{x}}^{{-{1}}}}}$

First, apply the negative exponent rule $\displaystyle{{a}}^{{-{m}}}=\frac{{1}}{{{a}}^{{m}}}$ .

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{6}}{{\frac{{1}}{{x}}+{2}}}+\frac{{3}}{{x}}}}$

Then, simplify the fraction $\displaystyle\frac{{6}}{{\frac{{1}}{{x}}+{2}}}$ . To do so, multiply it by $\displaystyle\frac{{x}}{{x}}$ .

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{6}}{{\frac{{1}}{{x}}+{2}}}\cdot\frac{{x}}{{x}}+\frac{{3}}{{x}}}}$

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{{6}{x}}}{{{1}+{2}{x}}}+\frac{{3}}{{x}}}}$

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{{6}{x}}}{{{2}{x}+{1}}}+\frac{{3}}{{x}}}}$

Then, multiply the top and bottom of the complex fraction by the LCD of the four fractions present. The LCD is $\displaystyle{{x}}^{{2}}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}$ .

$\displaystyle=\frac{{\frac{{3}}{{{x}}^{{2}}}+\frac{{1}}{{{2}{x}-{1}}}}}{{\frac{{{6}{x}}}{{{2}{x}+{1}}}+\frac{{3}}{{x}}}}\cdot\frac{{{{x}}^{{2}}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}}}{{{{x}}^{{2}}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}}}$

$\displaystyle=\frac{{{3}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}+{{x}}^{{2}}{\left({2}{x}+{1}\right)}}}{{{6}{x}{\left({{x}}^{{2}}\right)}{\left({2}{x}-{1}\right)}+{3}{x}{\left({2}{x}-{1}\right)}{\left({2}{x}+{1}\right)}}}$

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

$\displaystyle=\frac{{{\left({2}{x}+{1}\right)}{\left({3}{\left({2}{x}-{1}\right)}+{{x}}^{{2}}\right)}}}{{{3}{x}{\left({2}{x}-{1}\right)}{\left({2}{{x}}^{{2}}+{2}{x}+{1}\right)}}}$

$\displaystyle=\frac{{{\left({2}{x}+{1}\right)}{\left({6}{x}-{3}+{{x}}^{{2}}\right)}}}{{{3}{x}{\left({2}{x}-{1}\right)}{\left({2}{{x}}^{{2}}+{2}{x}+{1}\right)}}}$

$\displaystyle=\frac{{{\left({2}{x}+{1}\right)}{\left({{x}}^{{2}}+{6}{x}-{3}\right)}}}{{{3}{x}{\left({2}{x}-{1}\right)}{\left({2}{{x}}^{{2}}+{2}{x}+{1}\right)}}}$

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

$\displaystyle\frac{{{3}{{x}}^{{-{2}}}+{{\left({2}{x}-{1}\right)}}^{{-{1}}}}}{{\frac{{6}}{{{{x}}^{{-{1}}}+{2}}}+{3}{{x}}^{{-{1}}}}}=\frac{{{\left({2}{x}+{1}\right)}{\left({{x}}^{{2}}+{6}{x}-{3}\right)}}}{{{3}{x}{\left({2}{x}-{1}\right)}{\left({2}{{x}}^{{2}}+{2}{x}+{1}\right)}}}$ .

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