Question

How do you simplify ${{\left( -\dfrac{5{{f}^{9}}{{g}^{4}}{{h}^{2}}}{f{{g}^{2}}{{h}^{3}}} \right)}^{0}}$ ?

Answer 1

Hint: The given expression has terms that are in the exponential form in the numerator as well as the denominator. To simplify these exponents, we use the laws of exponents but first before we solve these exponents, we can see that there are brackets and power to them. According to the BODMAS rule, we need to first solve the brackets and the power first and then the contents inside. Following this evaluate the expression.

Complete step by step solution:
The given expression is, ${{\left( -\dfrac{5{{f}^{9}}{{g}^{4}}{{h}^{2}}}{f{{g}^{2}}{{h}^{3}}} \right)}^{0}}$
According to the BODMAS rule, the priority order of operators in solving any mathematical expression is,
B- Bracket
O- Order or power
D- Division
M- Multiplication
A- Addition
S- Subtraction
According to the BODMAS rule, we always solve the contents in the brackets first because it is of higher priority than the other mathematical operations.
So, let us solve the brackets first and the power to it.
Let us assume the contents inside it as x
Our expression will now be,
$\Rightarrow {{x}^{0}}$
According to the law of exponents,
Anything raised to the power of zero is equal to one.
Hence,
$\Rightarrow {{x}^{0}}=1$
Therefore, our expression, how big it might be, is raised to the power zero. Hence its value will be equal to one always.
$\Rightarrow {{\left( -\dfrac{5{{f}^{9}}{{g}^{4}}{{h}^{2}}}{f{{g}^{2}}{{h}^{3}}} \right)}^{0}}=1$

Note: We can prove this theory by the given following way.
The given law is that ${{x}^{0}}=1$
We can now write 0 as a number subtracted from itself.
$\Rightarrow {{x}^{a-a}}$
Now using the law of multiplication of exponents we can write,
$\Rightarrow {{x}^{a}}\times {{x}^{-a}}$
And upon further rewriting in the form,
$\Rightarrow {{x}^{a}}\times \dfrac{1}{{{x}^{a}}}$
Which on evaluation gives,
$\Rightarrow 1$
This is how this proof is apt.