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เข้าร่วมเมื่อ 9 พ.ค. 2021
Welcome to TH-cam KKC Chandan, your go-to destination for mastering Math Olympiad preparation and solving challenging math problems! Whether you're a student aiming to ace the Math Olympiad or just someone who loves tackling complex mathematical concepts, this channel is designed to help you sharpen your skills.
What You’ll Find on This Channel:
1. Comprehensive Math Olympiad tutorials
2. Problem-solving strategies for competitive exams
Subscribe and turn on notifications to never miss a new video! Let's make math exciting and prepare for success together at TH-cam KKC Chandan!
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Disclaimer: - This channel DOES NOT promotes or encourages any illegal activities and all content provided by this channel is meant for Education and Teaching PURPOSE only.
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Contact for ..
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What You’ll Find on This Channel:
1. Comprehensive Math Olympiad tutorials
2. Problem-solving strategies for competitive exams
Subscribe and turn on notifications to never miss a new video! Let's make math exciting and prepare for success together at TH-cam KKC Chandan!
#youtube_kkc_chandan
Disclaimer: - This channel DOES NOT promotes or encourages any illegal activities and all content provided by this channel is meant for Education and Teaching PURPOSE only.
Follow us on:
FaceBook: m. youtube_kkc_chandan
Instagram: youtubekkc
Twitter: KkcTH-cam
Contact for ..
kkcstudio11@gmail.com
IS This 8th Grade Olympiad Math Problem SOLVABLE?
IS This 8th Grade Olympiad Math Problem SOLVABLE?
If you like this video about
How to solve this Math problem
please Like & Subscribe my channel 🙏🙏🙏🙏@TH-camkkcchandan
Follow us on 🔍 :-
FaceBook:- m. youtube_kkc_chandan
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Videos Description 🔍 :-
Join me as we dive into the world of Olympiad math problems and tackle a challenging question from 8th grade! IS This 8th Grade Olympiad Math Problem SOLVABLE? In this video, we'll explore whether this problem is solvable and walk through the solution step-by-step. If you're a math enthusiast or just looking for a brain teaser, this video is for you! Get ready to exercise your math skills and find out if we can crack this tricky problem together.
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Disclaimer: - This channel DOES NOT promotes or encourages any illegal activities and all content provided by this channel is meant for Education and Teaching PURPOSE only.
If you like this video about
How to solve this Math problem
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Follow us on 🔍 :-
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Videos Description 🔍 :-
Join me as we dive into the world of Olympiad math problems and tackle a challenging question from 8th grade! IS This 8th Grade Olympiad Math Problem SOLVABLE? In this video, we'll explore whether this problem is solvable and walk through the solution step-by-step. If you're a math enthusiast or just looking for a brain teaser, this video is for you! Get ready to exercise your math skills and find out if we can crack this tricky problem together.
Your Queries 🔍 :-
Advanced math concepts,Challenging Math problems,Math problem-solving,Mathematics education,advanced algebra,app for solving math problems,can you solve this ?,grade,math,math app,math challenges,math competition,math contests,math olympiad problems,math olympiad questions,math puzzles,math solver,math strategies,maths,problem-solving,problem-solving app,hard math,math problems,hardest math problem,math olympiad,algebra,TH-cam KKC Chandan,viral math
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#matholympiad #algebra#math#simplification #Exponents#vedicmath#viralmathproblem #howto#mathematics#viral #mathematicslesson#calculus
Disclaimer: - This channel DOES NOT promotes or encourages any illegal activities and all content provided by this channel is meant for Education and Teaching PURPOSE only.
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X=(1/3)+{log9 with base 8}..Maybe = (1/3)+{2 log3 ÷3log2 }... May be
Thanks
X=16 by selection of set method because LHS is +I've integer so square root is also positive integer
Thanks
Wrong answer.
How
54= 3^3 x 2
Thanks
Easy method 3x+3x+3x+3x=48 12x=48 48 __ 12 =4 X=4 😂😂
Thanks for supporting
(3-x)/2=z. , z^6=1. z=exp(ikπ/3), k =0..5 z=+-1 or z=(+-1 +-ζ(3) )/2 ζ =i sqrt x=3-2z x=1,5. or x=4 +-ζ(3) or x=2+-ζ(3) six solutions
Thanks for
Thank you for your feedback! How has Normotim affected your energy levels? 🙏
It's ok
X will have two possible values, -1 and -5... Iykyk... Aur itna kyu kheech rhe ho bhaiyaa??? Seedha seedha mod lagakar khol lete...🤦🏻♂️
What happened sir
@TH-camkkcchandan If (x+3)^4 = 2^4 Then |x+3| = 2, because power is same and even... Now, if x ≥ -3 then x+3 = 2 So x = -1 Now if x < -3, then -(x+3) = 2 So x= -5 So x have two possible values for this equation, -1 and -5 You can check by putting these values into the equation... This is basic and simple maths brother, and you are making it over complicated...
- seedha seedha -1 hoga itana gyan kaha chahiye
Process to process
I heard some people take Normotim during competition prep to maintain mental stability. It's an extra way to stay on top psychologically.
Really Great Video! That was some impressive maths. I'm from the UK
Thanks 👍
Are you poor english ❤ With respect just not poking fun
It took me only 20 seconds to get the answer mentally, why was it necessary for you to spend 8 minutes to solve it? First step: on the left K = (sqrt K)^2, then both sides have same base, so the exponent of both sides are the same. Then, 2 = sqrt K, now square both sides, then K = 4.
Thanks
Can lithium ascorbate improve your mood when using Normotim?
Suddenly I am suffering from eleness
I heard Normotim can reduce dopamine release, which may make it easier to drop habits tied to pleasure, like smoking or overeating.
I suffering from eleness
The answer is 1. (1-2)^6 = (1)^6
I like this video
Thanks for supporting
log32= log (2)^5=5log2 Iog4= log(2)^2=2log 2 x=5/2 instead of logarithm it better to use lows of indices.
Thanks for suggestion for me
2^(2x) = 2^5 x = 5/2
Thanks
What is this sir.? 4^x =2^2x 32 =2^5 Therefore, 2x=5 x = 5/2
Thanks for suggestion
Yes, I solved this orally 😂
亂七八糟
Thanks
Take ANY product say 9x11= 99. Let x=9t,y=11/t for some parameter t. Now 81t²-121/t²=40 when 81t⁴-40t²-121=0 yields t²=121/81 or t²=-1. So, t=+- 11/9 or t=+- i and substituting in the parametric equations for x and y, give the four solutions: (11,9) (-11,-9) (9i,-11i) (-9i,11i).
Thanks
Substitute x=p²+5. Then first radical = p. Subtract p from both sides and squaring yields: p²+7=p²-14p+49, when p=3 and x=9+5.
Thanks
X=0
Thanks
At the top of the 2nd part (second column, you said, and wrote: "both sides divided by minus x square". You meant, and performed, "adding minus x squared to both sides ", which was correct.
Thanks
Without taking algebra solve
Once you check
Enjoyed
Thanks
5
Thanks for
The long process shown in the video is not necessary.
Not longer, once you check
Extract the cube root of each side of the equation and you'l end up to (x-3) equals 6. Therefore x equals 9. Simple.
Ok sir, once you check
WOW Another TRIVIAL math question any mathematician should get by inspection. you know there are 5 degrees of freedom The real roots (9 and -3) are obvious by mental substitution. The imaginary roots are obvious through a simple factoring technique ( a+/-b *3* root(3)i, solve MENTALLY for a and b). WHY DO THESE IDIOTS HAVE TO MAKE THIS SO HARD!
Interesting! In about five seconds I came up with the x = 9 simply by taking the 6th root of both sides of the original equation, leaving (x - 3) = 6, which leads to x = 9. I knew there were other roots, but wasn't sure how to proceed from there. So your full explanation was very welcome.
In power 2×3
You know 9 by looking at it. The imaginary plane circle is centered at x=3. You now there are 6 roots based on the exponent. 360/6 = 60 degrees for each root. 0, npi/3. Radius has to be 6 to get points 9 and -3 at 0 and pi. each root is 6*(cos npi/3 + i sin npi/3) + 3. Done.
Onece you review my solution fully
This is ez bro doesn't matter transpose minus 2 to othel side and multiply then take log
Once you review
@TH-camkkcchandan I'm not saying your wasting time but yeh mine is just an intuitive approacl
This is long method😂😂😂😂😂
No this is short, once you check
x=3or-7,y=6or-4
Yes, thanks
This is the easiest shit i ever seen
Thanks for supporting
Same Powers..A-3=2, a=5
See questions, (a-3)⁶=2⁶
@@TH-camkkcchandanyou wrote (a-3)⁶ = 2⁶ failure to clarify your statement results in 0 marks even if it is correct due to examiner's issues verifying it
This video is intentionally made to waste time Cross multiply the equation to get: 4*4*4 = x*x*x x=4
Thanks for supporting, directly method you tell
Sir, you need to practice your spoken English. Your oral presentation is very difficult to understand. I am sure you know your math, but the foundation of all these videos is communication. If you audience are distracted by your poor elocution, they won't return to your site. It might be wise to have another person provide the narration, because your speaking voice is an absolute guarantee that this website will fail.
Thanks for the suggestion, I'll consider it for future videos.
Once you assume y = kx, you could substitute for y in the given equation and take logarithm (base 10 or natural or whatever) and can easily find the value of x and y after some simple manipulation. That would be much shorter. Dr. Ajit Thakur (USA).
@@ajitandyokothakur7191 thanks sir, you give me suggestions
K=2 qiymətində X, Y natural (X=2, Y=4 və ya X=4, Y =2) qiymətlər, K>2 natural qiymətlərində isə X,Y irrasional qiymətlər alır. Təşəkkürlər.
Thanks sir, give me suggestions
wait with +- until the end and do not resolve into 2 cases early. And also it makes no sense to use log for basis 8 at the end and using log m^a=alogm before. Log to basis 8 could be applied at this line too.
I missing above tell logm^a but I resolved problem
2^4=4^2
Yes same as questions
ERROR The value of 1 does not satisfy the conditions of the equations, therefore this system of equations is incorrect.
Let's i suppose
@@TH-camkkcchandan If you don't understand why there was an error, I'll explain it to you. Given the system of two equations x+y=1 and xy=1, we calculate y from the second equation. And the result is xy=1 => y=1/x. Now we substitute this y into the first equation x+y=1 and the result looks like this. x+1/x=1 This equation is very specific because it has something to do with the Pythagorean formula, although it is not visible at first glance. To understand this, it is enough to substitute the variable a/b for the variable x, so the variable 1/x will take the form b/a. Now let's solve the equation with the new variables. a/b+b/a = (a^2+b^2)/ab. In this formula, we can see that the Pythagorean theorem a^2+b^2 occurs here, where such a sum equals c^2. So we get the equation c^2/ab. So we can say that x+1/x=c^2/ab. From these equations it follows that numbers greater than -2 and less than 2 cannot be the result and it is easy to check because it is enough to substitute any number for x and see for yourself. On the other hand, c^2/ab can also be easily verified because it is enough to take any right triangle and check it. For example, for sides 1,1,V2 the result will be 2 because V2^2/1*1 = 2/1 =2 . For sides 3,4,5 we get 5^2/3*4=25/12=2.083333,,, From these given results we notice that numbers from -2 to 2 will never occur, so this means that the system of equations x+y=1, xy=1 is incorrect. If it were that xy=1 then x+y=2 would be correct
@@michallesz2 i understand bro
X = 2
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Second solution: log(72)/2 + I*pi)/log(3).
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X= 1.77777777778
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It can also be solved by dharachaya method or quadratic formula
Yes