Question

Evaluate the value of $\dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{{x^{p - q}} + 1}}$ .

Answer 1

Hint: Here, we are asked to find the value of $\dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{{x^{p - q}} + 1}}$ .
Using the property ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$ , write ${x^{p - q}}$ in fraction form in the second term of the given sum of two fractions.
Then, take LCM as per the requirements and solve the sum further to get the required answer.

Complete step-by-step answer:
Here, we are asked to find the value of $\dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{{x^{p - q}} + 1}}$ .
Now, in the denominator of the second term of the given sum, there is ${x^{p - q}}$ .
Since, we know that, ${a^{m - n}}$ can be written as $\dfrac{{{a^m}}}{{{a^n}}}$ , i.e. ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$ .
So, using the above property of powers and exponents, we can write ${x^{p - q}}$ as $\dfrac{{{x^p}}}{{{x^q}}}$ .
Thus, $\dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{{x^{p - q}} + 1}} = \dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{\dfrac{{{x^p}}}{{{x^q}}} + 1}}$ .
Now, taking LCM in the denominator of the second term, we get
 $\dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{{x^{p - q}} + 1}} = \dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{\dfrac{{{x^p} + {x^q}}}{{{x^q}}}}}$
Also, $\dfrac{1}{{\dfrac{1}{a}}} = a$
 $\therefore \dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{{x^{p - q}} + 1}} = \dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{{{x^q}}}{{{x^p} + {x^q}}}$
Again, taking LCM in the above sum will give
 $\dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{{x^{p - q}} + 1}} = \dfrac{{{x^p} + {x^q}}}{{{x^p} + {x^q}}} = 1$
Thus, the value of $\dfrac{{{x^p}}}{{{x^p} + {x^q}}} + \dfrac{1}{{{x^{p - q}} + 1}}$ is 1.

Note: Some properties of powers and exponents are given as follows:
 $
  {a^m} \times {a^n} = {a^{m + n}} \\
  {a^n} \times {b^n} = {\left( {a \times b} \right)^n} \\
  \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}} \\
  \dfrac{{{a^n}}}{{{b^n}}} = {\left( {\dfrac{a}{b}} \right)^n} \\
  \dfrac{1}{{\dfrac{1}{a}}} = a \\
  {\left( {{a^m}} \right)^n} = {a^{m \times n}} \\
  \sqrt[n]{{{a^m}}} = {a^{\dfrac{m}{n}}} \\
  \sqrt[n]{a} = {a^{\dfrac{1}{n}}} \\
  {a^{ - n}} = \dfrac{1}{{{a^n}}} \\
  {a^0} = 1 \\
 $
Remember these properties.