Question

Find the value of the following:
i) $({3^0} + {4^{ - 1}}) \times {2^2}$
ii) $({2^{ - 1}} \times {4^{ - 1}}) \div {2^{ - 2}}$
iii) ${\left( {\dfrac{1}{2}} \right)^{ - 2}} + {\left( {\dfrac{1}{3}} \right)^{ - 2}} + {\left( {\dfrac{1}{4}} \right)^{ - 2}}$
iv) ${\left( {{3^{ - 1}} + {4^{ - 1}} + {5^{ - 1}}} \right)^0}$
v) ${\left( {{{\left( {\dfrac{{ - 2}}{3}} \right)}^{ - 2}}} \right)^2}$

Answer 1

Hint: Use the relation ${A^{ - x}} = \dfrac{1}{{{A^x}}}$ to convert the inverse terms into fractions. Use the BODMAS rule to prioritize the order of operations, where B-Brackets, O-Order, D-Division, M-Multiplication, A-Addition and S-Subtraction.

Complete step-by-step solution:
i) Given to us $({3^0} + {4^{ - 1}}) \times {2^2}$. We use the BODMAS rule to solve this expression.
So, according to the BODMAS rule, expressions inside the brackets should be solved first.
We know that any number/expression to the power is equal to $1$ and we can also write ${4^{ - 1}}$ as $\dfrac{1}{4}$
Now, the expression becomes $\left( {1 + \dfrac{1}{4}} \right) \times {2^2}$ . By solving this, we get $\left( {\dfrac{{4 + 1}}{4}} \right) \times 4 = \left( {\dfrac{5}{4}} \right) \times 4$
So the value of the given expression is $5$
ii) We have to solve $({2^{ - 1}} \times {4^{ - 1}}) \div {2^{ - 2}}$ . Firstly, we will convert the inverse values into fraction so the expression becomes $\left( {\dfrac{1}{2} \times \dfrac{1}{4}} \right) \div \dfrac{1}{{{2^2}}}$
Now, we solve the expression inside the brackets so that the expression becomes $\left( {\dfrac{1}{8}} \right) \div \dfrac{1}{{{2^2}}}$
Now this expression can be written as $\dfrac{{\left( {\dfrac{1}{8}} \right)}}{{\left( {\dfrac{1}{4}} \right)}}$ , by solving this we get the value $\dfrac{1}{2}$
Therefore, the value of the given expression is $\dfrac{1}{2}$
iii) To solve this, we have to remove the brackets first. We know that ${A^{ - x}} = \dfrac{1}{{{A^x}}}$ is also true vice verse so the given expression now becomes ${2^2} + {3^2} + {4^2}$
By solving, we get $4 + 9 + 16 = 29$
Hence the value of the given expression is $29$
iv) We already know that any number/expression to the power zero would be one. Therefore the value of the given expression is $1$
v) In the given expression ${\left( {{{\left( {\dfrac{{ - 2}}{3}} \right)}^{ - 2}}} \right)^2}$we have to remove the inverse first so we can write this expression as ${\left( {{{\left( {\dfrac{{ - 3}}{2}} \right)}^2}} \right)^2}$
By solving, we get ${\left( {\dfrac{9}{4}} \right)^2}$
Therefore, the value of the given expression is $\dfrac{{81}}{{16}}$

Note: One might mistake the value of any expression or number to the power zero would be zero, but the value is $1$ and not zero. Note that the BODMAS rule applies for every type of equation/expression and they should only be solved satisfying these rules.