
Find and correct errors of the following mathematical expressions:
$ \dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + 1 = 2 $
Answer
508.8k+ views
Hint: The numerator can break into two parts, if they are sharing the same denominator
i.e. $ \dfrac{{A + B}}{C} = \dfrac{A}{C} + \dfrac{B}{C} $
And then compare the values of LHS terms and RHS terms to find out the errors.
Complete step-by-step answer:
1. We are provided with the expression:
$ \dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + 1 = 2 $
Which is wrong.
2. Considering the L.H.S of the expression:
$ \dfrac{{3{x^2} + 1}}{{3{x^2}}} $
Using the hint provided above we can say that
$ \begin{gathered}
\dfrac{{3{x^2} + 1}}{{3{x^2}}} = \dfrac{{3{x^2}}}{{3{x^2}}} + \dfrac{1}{{3{x^2}}} \\
\\
\end{gathered} $
Since the first terms are same in both numerator and denominator, we get
= $ 1 + \dfrac{1}{{3{x^2}}} $
Therefore the correct answer of the expression is:
$ \dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + \dfrac{1}{{3{x^2}}} $
Therefore, L.H.S=R.H.S.
Note: We can also add two rational numbers whose denominators are not the same by multiplying their numerators and denominators by an appropriate number so that the denominators can match and can have a common denominator.
i.e. $ \dfrac{{A + B}}{C} = \dfrac{A}{C} + \dfrac{B}{C} $
And then compare the values of LHS terms and RHS terms to find out the errors.
Complete step-by-step answer:
1. We are provided with the expression:
$ \dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + 1 = 2 $
Which is wrong.
2. Considering the L.H.S of the expression:
$ \dfrac{{3{x^2} + 1}}{{3{x^2}}} $
Using the hint provided above we can say that
$ \begin{gathered}
\dfrac{{3{x^2} + 1}}{{3{x^2}}} = \dfrac{{3{x^2}}}{{3{x^2}}} + \dfrac{1}{{3{x^2}}} \\
\\
\end{gathered} $
Since the first terms are same in both numerator and denominator, we get
= $ 1 + \dfrac{1}{{3{x^2}}} $
Therefore the correct answer of the expression is:
$ \dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + \dfrac{1}{{3{x^2}}} $
Therefore, L.H.S=R.H.S.
Note: We can also add two rational numbers whose denominators are not the same by multiplying their numerators and denominators by an appropriate number so that the denominators can match and can have a common denominator.
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