Olympiad Mathematics | Learn how to solve this problem | Olympiad Training

[@mathbynisharsir5586]

Amazing explanation sir

[@PreMath]

Thanks and welcome You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@MK06086]

I am obliged to say as an 2nd grade of high school student that I am amazed that the math community exists and is thriving in answering this types of problems! Well done sir I wish you the best of all wishes

[@rubensramos6458]

At a first glance, it seems to be difficult. However, the teacher makes it soft and easy.

[@ghhdcdvv5069]

تمرين جيد جميل .شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة فلسطين.

[@PreMath]

Thanks dear🙏

[@garypaulson5202]

Thank you professor, these videos are fun to watch (and sometimes catch myself making mistakes!)

[@AdityaKumar-gv4dj]

x=2 and x=-2 finally I am able to solve these type of famous sums

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@e-learningnetwork]

kzbin.info/door/hpvgXVuXCY26LSF6Dh0rTA

[@abodamohye3195]

I love this channel so much,I hope you come forward❤️

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!!

[@PreMath]

So nice of you. Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@d.sambasivaraosivarao2852]

Sir please explain any topic like integration,differentiation,etc...

[@PreMath]

Keep watching... Great suggestion. Cheers

[@dreamaker2107]

Can you please solve this in a following video: Solve the system for all the real numbers: 1/x + 1/y = 1 y/x² + x/y² = 1 (It was on a math competition that I took part)

[@ib5299]

According to me both y and x are 2 . I am not too good on math so don't take this answer seriously

[@dreamaker2107]

@@ib5299 Yeah I know the answer too... (It's not difficult to figure it out). I want to see how he will solve it

[@ib5299]

@@dreamaker2107 I solved it by squaring the first equation then etc

[@PreMath]

Please send your question to the email: [e-mail] Thanks for asking. Cheers

[@ib5299]

@@PreMath I emailed you a problem can you please solve it , it is really tricky.

[@SuperYoonHo]

thanks sir

[@nirupamasingh2948]

Vv nice presentation

[@PreMath]

Thank you! Cheers! You are awesome, Niru. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@nirupamasingh2948]

@@PreMath 🙏🙏

[@TiêuDaoTiên23]

amazing

[@pranavamali05]

Thnku

[@PreMath]

You are very welcome! You are awesome, Pranav. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@pramodsingh7898]

Thanks

[@PreMath]

You are very welcome! You are awesome, Pramod. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@ahmedam8187]

That’s nice

[@susmitasannal2539]

Where do I get more such types of sums to practice

[@swastikaacherjee4797]

Please tell me too

[@lemonade9913]

Thank u sir

[@PreMath]

You are very welcome! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@msafasharhan]

Hello Sir ,Is there another methode to solve this problem because this solution i can't under stand

[@mehulpunia6174]

Very nice

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@petrileskinen2988]

My solution involving the inverse hyperbolic cosine: First substitute a = sqrt(2+sqrt(3)) now, knowing that a^(-1) = sqrt(2-sqrt(3)) the equation becomes a^x + a^(-x) = 4 (e^(x ln a) + e^(-x ln a))/2 = 2 cosh (x ln a) = 2 x ln a = arcosh (2) x = arcosh (2) / ln a substitute arcosh(z) with ± ln(z+sqrt(z^2-1)) x = ± ln(2+sqrt(2^2-1) / ln sqrt(2+sqrt(3)) = ± ln(2+sqrt(2^2-1) / (1/2 ln(2+sqrt(2^2-1)) = ... etc ...

[@rubensramos6458]

The general solution for this kind of problem can be found in "Solving the Fractional Polynomial ax^r +bx^s +c = 0 Using the Lambert-Tsallis Wq Function", that is availiable for download on Researchgate. The solutions are x1=(1/log(b/a))*log((1/z)*Wq(z*(c^z))) = 2 (q = 1-1/z) and x2=(1/log(a/b))*log((1/y)*Wq(y*(c^y))) = -2 (q=1-1/y). a = sqrt(2+sqrt(3)), b = sqrt(2-sqrt(3)), z =log(b/a)/log(a) and y=log(a/b)/log(b).

[@randomjudgements8852]

Very simple 2and -2

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@giuseppemalaguti435]

Pongo t=(sqrt (2+sqrt 3))^x...ottengo t+1/t=4,t^2-4t+1=0,t=2+sqrt 3,t=2-sqrt3...risulta x=2,x=-2

[@vacuumcarexpo]

（2cos 15°）^x＋（2sin 15°）^x＝4

[@에스피-h8t]

Solution by insight x=2 orx=-2

[@ujwalsmanhas1093]

easy question....

[@shahedshahrear3918]

x ^x/2 + y ^x/2

[@rudychan8792]

Looks Complicated. But, we can solve this by Guessing^ 😉 x = 0 --> f(x) = 1 + 1 = 2 x = 1 --> f(x) = √a + √b = √6 Then, magic, bingo: x = 2 --> f(x) = 2+√3 + 2-√3 = 4 Easy Peasy^ 😉 😁

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@ruilongsheng2845]

x=±2

[@nsnn980]

For kids :)