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Math Olympiad 9^x-6^x=4^x😊| Math Olympiad Problems | Algebra @Mamtamaam
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Germany | Can you solve this? | A Nice Math olympiad Algebra Promblem
Math Olympiad 3^m-2^m=65 | Math Olympiad Problems|Algebra
Math Olympiad 3^m-2^m=65
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A Very Nice Maths Olympiad Question | How to solve (4 + \sqrt{5}^x+ (4-\sqrt{5})^x | Algebra | 🤔
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• Mummy Ki Aisi Demand.....
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Maths Olympaid are held all around the world to recognise students who excel in maths. Thanks for Watching .
Maths Olympaid are held all around the world to recognise students who excel in maths.
Thanks for Watching .

[@harshpathak2359]

Nice solution

[@manojchaugule794]

nice explaination

[@user-go5sy7mu6q]

Very good explanation

[@user-tc9cb8zh7b]

True math class.Good one

[@arvindkulkarni6580]

Perfect Teaching Method

[@aruntiwari5714]

Very well done maam

[@dhananjaypandey6258]

Lovely explanation 👌

[@harikishan1900]

It's very well explained here

[@LaxmanKumar-yi4cv]

It's very effective way to learn and understand

[@Maheshkumar-cw6un]

Hello mam,its good question

[@satyadev3848]

its pure detail level of explaination here

[@professorsargeanthikesclim9293]

Multiply both sides by 16 = 2^4. (2 * x)^4 = (2 * (x - 4))^4 = (2x - 8)^4. Let u the arithmetic mean of 2x and 2x - 8; i.e., u = 2x - 4. So, 2x = u +4 and 2x - 8 = (u + 4) - 8 = u - 4. Thus, (u + 4)^4 = (u + 4)^4 u^4 + 16 u^3 + 96 u^2 + 256 u + 256 = u^4 - 16 u^3 + 96 u^2 - 256 u + 256. 0= 32 * u^3 + 512 * u = 32 * u * (u^2 + 16) = 32 * u * (u + 4i)(u - 4i). Divide both sides by 32. 2x - 4 = u = 0, 4i or -4i. x = 2, 2 + 2i or 2 - 2i.

[@ritwikgupta3655]

Q is X=(X-4)^4, pls note. LHS is only X, not X^4 as you write.

[@DanDart]

What you wrote first and your thumbnail are mismatched

[@Robert-er5wq]

I think the equation is x = (x-4)^4 and not x^4=(x-4)^4