Question

2^2^2-[(2^2)^2]^2
Try if u can solve it.
(The question is not wrong) ​

Answer 1

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Given :

$\sf \displaystyle{\Big(\big(2\big)^2\Big)^2 -\Big[ (2^2)^2 \Big ]^2}$

Solution :

• Solve the Inner brackets initialy.
• Solve the powers one by one 9

$\sf \implies \displaystyle{\Big(\big(2\big)^2\Big)^2 -\Big[ (2^2)^2 \Big ]^2}$

$\implies \displaystyle {\big(4\big)^2}-\big[(4)^2\big]^2$

$\sf \implies 16 - \big[16\big]^2$

$\sf \implies 16 - 256$

$\sf \implies -240$

Required Answer :

The value of ${\sf \Big(\big(2\big)^2\Big)^2 -\Big[ (2^2)^2 \Big ]^2 }$ is $\underline {\sf -240}$

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$\underline {\large{\sf Maths \:Flash :}}$

• 9 is also known as the magic number. This is because if you multiply a number by 9 and add all the digits of the new number together, the sum will always add up to 9. For example: 8 x 9 = 72 or, 7 + 2 = 9
• Two and five are the only prime numbers in the entire number system which ends with two or five.
• If the sum of the digits is divisible by three, then the given number will also be divisible by three. This is known as the rule of divisibility of three.

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Answer 2

Answer:

refer to the attachment!

Step-by-step explanation:

hope it helps you!!:)

SARANGHAE!