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Write the following as simply as possible without using negative indices and as a single fraction: 3-2x⁻²(xy)(1)/(2) + (1)/(x)

Mathematics

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Answer

c

1+9x32y129x52y12\frac{1 + 9x^{\frac{3}{2}}y^{\frac{1}{2}}}{9x^{\frac{5}{2}}y^{\frac{1}{2}}}

Step-by-step solution

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Part 4

Write the following as simply as possible without using negative indices and as a single fraction.

Part c

1+9x32y129x52y12\frac{1 + 9x^{\frac{3}{2}}y^{\frac{1}{2}}}{9x^{\frac{5}{2}}y^{\frac{1}{2}}}

3−2x−2(xy)12+1x\frac{3^{-2}x^{-2}}{(xy)^{\frac{1}{2}}} + \frac{1}{x}

  1. Convert terms with negative exponents to positive exponents.

    3−2=132=193^{-2} = \frac{1}{3^2} = \frac{1}{9} x−2=1x2x^{-2} = \frac{1}{x^2}

    The numerator of the first term becomes 19x2\frac{1}{9x^2}.

  2. Apply the exponent rule (ab)n=anbn(ab)^n = a^n b^n to the denominator of the first term.

    (xy)12=x12y12(xy)^{\frac{1}{2}} = x^{\frac{1}{2}}y^{\frac{1}{2}}
  3. Rewrite the first fraction by combining the results from Step 1 and Step 2.

    19x2x12y12=19x2x12y12\frac{\frac{1}{9x^2}}{x^{\frac{1}{2}}y^{\frac{1}{2}}} = \frac{1}{9x^2 x^{\frac{1}{2}}y^{\frac{1}{2}}}
  4. Combine the xx terms in the denominator of the first fraction using the rule am⋅an=am+na^m \cdot a^n = a^{m+n}.

    x2⋅x12=x2+12=x42+12=x52x^2 \cdot x^{\frac{1}{2}} = x^{2 + \frac{1}{2}} = x^{\frac{4}{2} + \frac{1}{2}} = x^{\frac{5}{2}}

    So the first fraction is 19x52y12\frac{1}{9x^{\frac{5}{2}}y^{\frac{1}{2}}}.

  5. The expression is now 19x52y12+1x\frac{1}{9x^{\frac{5}{2}}y^{\frac{1}{2}}} + \frac{1}{x}.

    To combine these into a single fraction, find the least common denominator (LCD). The LCD for 9x52y129x^{\frac{5}{2}}y^{\frac{1}{2}} and xx is 9x52y129x^{\frac{5}{2}}y^{\frac{1}{2}}.

  6. Adjust the second term to have the LCD.

    To change the denominator xx to 9x52y129x^{\frac{5}{2}}y^{\frac{1}{2}}, we multiply the numerator and denominator by 9x32y129x^{\frac{3}{2}}y^{\frac{1}{2}}.

    1x=1⋅(9x32y12)x⋅(9x32y12)=9x32y129x1+32y12=9x32y129x52y12\frac{1}{x} = \frac{1 \cdot (9x^{\frac{3}{2}}y^{\frac{1}{2}})}{x \cdot (9x^{\frac{3}{2}}y^{\frac{1}{2}})} = \frac{9x^{\frac{3}{2}}y^{\frac{1}{2}}}{9x^{1+\frac{3}{2}}y^{\frac{1}{2}}} = \frac{9x^{\frac{3}{2}}y^{\frac{1}{2}}}{9x^{\frac{5}{2}}y^{\frac{1}{2}}}
  7. Add the two fractions.

    19x52y12+9x32y129x52y12=1+9x32y129x52y12\frac{1}{9x^{\frac{5}{2}}y^{\frac{1}{2}}} + \frac{9x^{\frac{3}{2}}y^{\frac{1}{2}}}{9x^{\frac{5}{2}}y^{\frac{1}{2}}} = \frac{1 + 9x^{\frac{3}{2}}y^{\frac{1}{2}}}{9x^{\frac{5}{2}}y^{\frac{1}{2}}}

    The final answer is 1+9x32y129x52y12\boxed{\frac{1 + 9x^{\frac{3}{2}}y^{\frac{1}{2}}}{9x^{\frac{5}{2}}y^{\frac{1}{2}}}}.

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