
How do you simplify $ \dfrac{{{a^{ - 1}} + {b^{ - 1}}}}{{{a^{ - 2}} + {b^{ - 2}}}}? $
Answer
520.5k+ views
Hint: To simplify the given expression, you have to use the law of indices for negative powers to simplify the individual terms with negative powers into positive powers then take LCM and add accordingly. And then after seeing if there is any common term in the numerator and the denominator to be cancelled out, if yes then cancel it otherwise write the simplified expression.
Complete step by step solution:
In order to simplify the given expression $ \dfrac{{{a^{ - 1}} + {b^{ - 1}}}}{{{a^{ - 2}} + {b^{ - 2}}}} $ we will go through the law of indices for negative power, which says if a term has a negative power it can expressed as the multiplicative inverse of the term with positive power as follows
$ {x^{ - y}} = \dfrac{1}{{{x^y}}} $
Using the law of indices for negative powers, we will get
$ \dfrac{{{a^{ - 1}} + {b^{ - 1}}}}{{{a^{ - 2}} + {b^{ - 2}}}} = \dfrac{{\dfrac{1}{{{a^1}}} + \dfrac{1}{{{b^1}}}}}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}} $
We can also write it as
$ \dfrac{{\dfrac{1}{{{a^1}}} + \dfrac{1}{{{b^1}}}}}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}} = \dfrac{{\dfrac{1}{a} + \dfrac{1}{b}}}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}} $
Now, taking Lowest Common Factor to add them, we will get
$ \dfrac{{\dfrac{1}{a} + \dfrac{1}{b}}}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}} = \dfrac{{\dfrac{{b + a}}{{ab}}}}{{\dfrac{{{b^2} + {a^2}}}{{{a^2}{b^2}}}}} $
We can see that numerator and denominator part are themselves become fraction, and also we know that a fraction is nothing but a way of writing numerator being divided by dominator, so we will use division sign here to express this fraction as follows
$ \dfrac{{\dfrac{{b + a}}{{ab}}}}{{\dfrac{{{b^2} + {a^2}}}{{{a^2}{b^2}}}}} = \dfrac{{b + a}}{{ab}} \div \dfrac{{{b^2} + {a^2}}}{{{a^2}{b^2}}} $
Since multiplication and division are inverse function to each other so if we inverse the second fraction then sign will be changed to multiplication,
$ \dfrac{{b + a}}{{ab}} \div \dfrac{{{b^2} + {a^2}}}{{{a^2}{b^2}}} = \dfrac{{b + a}}{{ab}} \times \dfrac{{{a^2}{b^2}}}{{{b^2} + {a^2}}} $
Simplifying it further,
$ \dfrac{{b + a}}{{ab}} \times \dfrac{{{a^2}{b^2}}}{{{b^2} + {a^2}}} = \dfrac{{\left( {b + a} \right)ab}}{{\left( {{b^2} + {a^2}} \right)}} $
Therefore $ \dfrac{{\left( {b + a} \right)ab}}{{\left( {{b^2} + {a^2}} \right)}} $ is the simplified form of the given expression.
So, the correct answer is “$ \dfrac{{\left( {b + a} \right)ab}}{{\left( {{b^2} + {a^2}} \right)}} $”.
Note: When writing a fraction into division notation then take care of the fact that the numerator should be written on the left side of the division sign whereas the denominator on the right side, because the numerator is being divided by the denominator.
Complete step by step solution:
In order to simplify the given expression $ \dfrac{{{a^{ - 1}} + {b^{ - 1}}}}{{{a^{ - 2}} + {b^{ - 2}}}} $ we will go through the law of indices for negative power, which says if a term has a negative power it can expressed as the multiplicative inverse of the term with positive power as follows
$ {x^{ - y}} = \dfrac{1}{{{x^y}}} $
Using the law of indices for negative powers, we will get
$ \dfrac{{{a^{ - 1}} + {b^{ - 1}}}}{{{a^{ - 2}} + {b^{ - 2}}}} = \dfrac{{\dfrac{1}{{{a^1}}} + \dfrac{1}{{{b^1}}}}}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}} $
We can also write it as
$ \dfrac{{\dfrac{1}{{{a^1}}} + \dfrac{1}{{{b^1}}}}}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}} = \dfrac{{\dfrac{1}{a} + \dfrac{1}{b}}}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}} $
Now, taking Lowest Common Factor to add them, we will get
$ \dfrac{{\dfrac{1}{a} + \dfrac{1}{b}}}{{\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}}} = \dfrac{{\dfrac{{b + a}}{{ab}}}}{{\dfrac{{{b^2} + {a^2}}}{{{a^2}{b^2}}}}} $
We can see that numerator and denominator part are themselves become fraction, and also we know that a fraction is nothing but a way of writing numerator being divided by dominator, so we will use division sign here to express this fraction as follows
$ \dfrac{{\dfrac{{b + a}}{{ab}}}}{{\dfrac{{{b^2} + {a^2}}}{{{a^2}{b^2}}}}} = \dfrac{{b + a}}{{ab}} \div \dfrac{{{b^2} + {a^2}}}{{{a^2}{b^2}}} $
Since multiplication and division are inverse function to each other so if we inverse the second fraction then sign will be changed to multiplication,
$ \dfrac{{b + a}}{{ab}} \div \dfrac{{{b^2} + {a^2}}}{{{a^2}{b^2}}} = \dfrac{{b + a}}{{ab}} \times \dfrac{{{a^2}{b^2}}}{{{b^2} + {a^2}}} $
Simplifying it further,
$ \dfrac{{b + a}}{{ab}} \times \dfrac{{{a^2}{b^2}}}{{{b^2} + {a^2}}} = \dfrac{{\left( {b + a} \right)ab}}{{\left( {{b^2} + {a^2}} \right)}} $
Therefore $ \dfrac{{\left( {b + a} \right)ab}}{{\left( {{b^2} + {a^2}} \right)}} $ is the simplified form of the given expression.
So, the correct answer is “$ \dfrac{{\left( {b + a} \right)ab}}{{\left( {{b^2} + {a^2}} \right)}} $”.
Note: When writing a fraction into division notation then take care of the fact that the numerator should be written on the left side of the division sign whereas the denominator on the right side, because the numerator is being divided by the denominator.
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