
Simplify
i) \[\left( {{3^0} - {2^0}} \right) \times \left( {{3^0} + {2^0}} \right)\]
ii) \[\left( {{5^{15}} \div {5^{10}}} \right) \times {5^5}\]
Answer
502.2k+ views
Hint: The question is related to the exponential numbers. Here we have to simplify the both terms, so by using the laws of indices which relates to the exponential number we are simplifying the both terms. Hence it will be a required solution for the given question.
Complete step-by-step answer:
The numbers are in the form of the exponential number, It is defined as a number of times the number is multiplied by itself. When we multiply the exponential number we follow the laws of indices. Between the exponential number there is an arithmetic operation we will apply that operations for the terms.
The laws of indices are given as
\[{a^m} \times {a^n} = {a^{m + n}}\]
\[{a^m} \div {a^n} = {a^{m - n}}\]
\[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[{a^0} = 1\]
\[{a^{ - 1}} = \dfrac{1}{a}\]
Now on considering given question
i) \[\left( {{3^0} - {2^0}} \right) \times \left( {{3^0} + {2^0}} \right)\]
First we simplify the terms which is present in the braces
By the law of the indices we know that \[{a^0} = 1\], now the given equation is written as
\[ \Rightarrow \left( {1 - 1} \right) \times \left( {1 + 1} \right)\]
On simplifying we have
\[ \Rightarrow \left( 0 \right) \times \left( 2 \right)\]
Any number multiplied by the zero then the product will be zero. So we have
\[ \Rightarrow 0\]
Therefore \[\left( {{3^0} - {2^0}} \right) \times \left( {{3^0} + {2^0}} \right) = 0\]
ii) \[\left( {{5^{15}} \div {5^{10}}} \right) \times {5^5}\]
First we simplify the terms which is present in the braces
By the law of the indices we know that \[{a^m} \div {a^n} = {a^{m - n}}\], now the given equation is written as
\[ \Rightarrow \left( {{5^{15 - 10}}} \right) \times {5^5}\]
On simplifying we have
\[ \Rightarrow \left( {{5^5}} \right) \times \left( {{5^5}} \right)\]
By the law of the indices we know that \[{a^m} \times {a^n} = {a^{m + n}}\], now the given equation is written as
\[ \Rightarrow {5^{5 + 5}}\]
On simplifying we have
\[ \Rightarrow {5^{10}}\]
Therefore \[\left( {{5^{15}} \div {5^{10}}} \right) \times {5^5} = {5^{10}}\]
Hence simplified
Note: In mathematics we have four different kinds of arithmetic operations namely, addition, subtraction, multiplication and division. When there is more than one arithmetic operation then we are following the BODMAS rule which is known as Bracket Of Division Multiplication Addition Subtraction.
Complete step-by-step answer:
The numbers are in the form of the exponential number, It is defined as a number of times the number is multiplied by itself. When we multiply the exponential number we follow the laws of indices. Between the exponential number there is an arithmetic operation we will apply that operations for the terms.
The laws of indices are given as
\[{a^m} \times {a^n} = {a^{m + n}}\]
\[{a^m} \div {a^n} = {a^{m - n}}\]
\[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[{a^0} = 1\]
\[{a^{ - 1}} = \dfrac{1}{a}\]
Now on considering given question
i) \[\left( {{3^0} - {2^0}} \right) \times \left( {{3^0} + {2^0}} \right)\]
First we simplify the terms which is present in the braces
By the law of the indices we know that \[{a^0} = 1\], now the given equation is written as
\[ \Rightarrow \left( {1 - 1} \right) \times \left( {1 + 1} \right)\]
On simplifying we have
\[ \Rightarrow \left( 0 \right) \times \left( 2 \right)\]
Any number multiplied by the zero then the product will be zero. So we have
\[ \Rightarrow 0\]
Therefore \[\left( {{3^0} - {2^0}} \right) \times \left( {{3^0} + {2^0}} \right) = 0\]
ii) \[\left( {{5^{15}} \div {5^{10}}} \right) \times {5^5}\]
First we simplify the terms which is present in the braces
By the law of the indices we know that \[{a^m} \div {a^n} = {a^{m - n}}\], now the given equation is written as
\[ \Rightarrow \left( {{5^{15 - 10}}} \right) \times {5^5}\]
On simplifying we have
\[ \Rightarrow \left( {{5^5}} \right) \times \left( {{5^5}} \right)\]
By the law of the indices we know that \[{a^m} \times {a^n} = {a^{m + n}}\], now the given equation is written as
\[ \Rightarrow {5^{5 + 5}}\]
On simplifying we have
\[ \Rightarrow {5^{10}}\]
Therefore \[\left( {{5^{15}} \div {5^{10}}} \right) \times {5^5} = {5^{10}}\]
Hence simplified
Note: In mathematics we have four different kinds of arithmetic operations namely, addition, subtraction, multiplication and division. When there is more than one arithmetic operation then we are following the BODMAS rule which is known as Bracket Of Division Multiplication Addition Subtraction.
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