Answer
Verified
349.2k+ views
Hint: The question is in the form of exponential. It is in the form of \[\left( {{{\left( {{a^2}} \right)}^2} + {b^2}} \right)\], The b term for each and every term is same. The value of a is different for other terms. The given terms are in the form of \[({x^4} + 324)\], this can be written as \[({(x + 3)^2} + 9)({(x - 3)^2} + 9)\], this formula is implemented to each term and then we are going to simplify.
Complete step by step answer:
Consider the term \[({(x + 3)^2} + 9)({(x - 3)^2} + 9)\], on multiplying this we have
\[ \Rightarrow {(x + 3)^2}({(x - 3)^2} + 9) + 9({(x - 3)^2} + 9)\]
Multiplying to each term we have
\[ \Rightarrow {(x + 3)^2}{(x - 3)^2} + 9{(x + 3)^2} + 9{(x - 3)^2} + 81\]
On simplifying we have
\[ \Rightarrow {({x^2} - 9)^2} + 9(2{x^2} + 18) + 81\]
Expanding the terms
\[ \Rightarrow {x^4} + 81 - 18{x^2} + 18{x^2} + 162 + 81\]
On further simplification we have
\[ \Rightarrow {x^4} + 324\]
Therefore \[({x^4} + 324) = ({(x - 3)^2} + 9)({(x + 3)^2} + 9)\]----------- (1)
Now consider the given question
\[ \Rightarrow \dfrac{{({{10}^4} + 324)({{22}^4} + 324)({{34}^4} + 324)({{46}^4} + 324)({{58}^4} + 324)}}{{({4^4} + 324)({{16}^4} + 324)({{28}^4} + 324)({{40}^4} + 324)({{52}^4} + 324)}}\]
By using the equation (1) the above inequality is written as
\[ \Rightarrow \dfrac{{({7^2} + 9)({{13}^2} + 9)({{19}^2} + 9)({{25}^2} + 9)({{31}^2} + 9)({{37}^2} + 9)({{43}^2} + 9)({{49}^2} + 9)({{55}^2} + 9)({{61}^2} + 9)}}{{({1^2} + 9)({7^2} + 9)({{13}^2} + 9)({{19}^2} + 9)({{25}^2} + 9)({{31}^2} + 9)({{37}^2} + 9)({{43}^2} + 9)({{49}^2} + 9)({{55}^2} + 9)}}\]
When we see both the numerator and the denominator there are so many common terms hence we can cancel them.
On cancelling them we have
\[ \Rightarrow \dfrac{{({{61}^2} + 9)}}{{({1^2} + 9)}}\]
Now we have a very simplified term, in the numerator there is only one term and in the denominator also we can see only one term.
The terms which are in the form of exponents. On expanding it we have
\[ \Rightarrow \dfrac{{(61 \times 61 + 9)}}{{(1 \times 1 + 9)}}\]
Now multiply 61 and 61 in the numerator. Multiply 1 and 1 in the denominator.
\[ \Rightarrow \dfrac{{(3721 + 9)}}{{(1 + 9)}}\]
Now add 3721 and 9 in the numerator. Add 1 and 9 in the denominator.
\[ \Rightarrow \dfrac{{3730}}{{10}}\]
On dividing the 3730 by 10
\[ \Rightarrow 373\]
Hence we have simplified the given inequality.
Note: The number can be written in the form of the exponents and the exponent number can also be written in the form of the usual number. We must know about the law of exponents. Since the given terms are exponential, we need a law of exponents to solve.
Complete step by step answer:
Consider the term \[({(x + 3)^2} + 9)({(x - 3)^2} + 9)\], on multiplying this we have
\[ \Rightarrow {(x + 3)^2}({(x - 3)^2} + 9) + 9({(x - 3)^2} + 9)\]
Multiplying to each term we have
\[ \Rightarrow {(x + 3)^2}{(x - 3)^2} + 9{(x + 3)^2} + 9{(x - 3)^2} + 81\]
On simplifying we have
\[ \Rightarrow {({x^2} - 9)^2} + 9(2{x^2} + 18) + 81\]
Expanding the terms
\[ \Rightarrow {x^4} + 81 - 18{x^2} + 18{x^2} + 162 + 81\]
On further simplification we have
\[ \Rightarrow {x^4} + 324\]
Therefore \[({x^4} + 324) = ({(x - 3)^2} + 9)({(x + 3)^2} + 9)\]----------- (1)
Now consider the given question
\[ \Rightarrow \dfrac{{({{10}^4} + 324)({{22}^4} + 324)({{34}^4} + 324)({{46}^4} + 324)({{58}^4} + 324)}}{{({4^4} + 324)({{16}^4} + 324)({{28}^4} + 324)({{40}^4} + 324)({{52}^4} + 324)}}\]
By using the equation (1) the above inequality is written as
\[ \Rightarrow \dfrac{{({7^2} + 9)({{13}^2} + 9)({{19}^2} + 9)({{25}^2} + 9)({{31}^2} + 9)({{37}^2} + 9)({{43}^2} + 9)({{49}^2} + 9)({{55}^2} + 9)({{61}^2} + 9)}}{{({1^2} + 9)({7^2} + 9)({{13}^2} + 9)({{19}^2} + 9)({{25}^2} + 9)({{31}^2} + 9)({{37}^2} + 9)({{43}^2} + 9)({{49}^2} + 9)({{55}^2} + 9)}}\]
When we see both the numerator and the denominator there are so many common terms hence we can cancel them.
On cancelling them we have
\[ \Rightarrow \dfrac{{({{61}^2} + 9)}}{{({1^2} + 9)}}\]
Now we have a very simplified term, in the numerator there is only one term and in the denominator also we can see only one term.
The terms which are in the form of exponents. On expanding it we have
\[ \Rightarrow \dfrac{{(61 \times 61 + 9)}}{{(1 \times 1 + 9)}}\]
Now multiply 61 and 61 in the numerator. Multiply 1 and 1 in the denominator.
\[ \Rightarrow \dfrac{{(3721 + 9)}}{{(1 + 9)}}\]
Now add 3721 and 9 in the numerator. Add 1 and 9 in the denominator.
\[ \Rightarrow \dfrac{{3730}}{{10}}\]
On dividing the 3730 by 10
\[ \Rightarrow 373\]
Hence we have simplified the given inequality.
Note: The number can be written in the form of the exponents and the exponent number can also be written in the form of the usual number. We must know about the law of exponents. Since the given terms are exponential, we need a law of exponents to solve.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE