Simplification Method for Math Olympiad | An Algebra Challenge
Simplification Method for Math Olympiad | An Algebra Challenge
Unlock the secrets to mastering simplification strategies for Math Olympiad success! In this video, we delve into essential algebraic techniques that will help you simplify complex problems with ease. Whether you're a Math Olympiad participant or just looking to sharpen your algebra skills, this challenge is designed to enhance your problem-solving abilities. Join us as we break down intricate problems and reveal tips and tricks to make simplification second nature. Are you ready to tackle this algebraic challenge?
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🔢 What You'll Learn:
Key strategies for simplifying complex expressions
Tips and tricks to approach difficult simplification problems
Step-by-step walkthrough of the solution
🧠 Challenge Yourself:
Pause the video, try to solve the problem on your own, and then watch as we break down the solution. Share your approach and answers in the comments below!
Timestamps:
0:00 Introduction
2:55 Algebraic manipulations
6:32 Simplifying Expression
8:22 Evaluating Expression
10:30 Answer
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Thanks for Watching !
Пікірлер: 8
A wonderful introduction....thanks for sharing Sir 🙏...x= 1/( 77)^99 ..
0+1/(77)^99= 1/ (77)^99 soln
1/(77)^99
Silly to use all these manipulations to solve something that you can deduce through inspection. The two equations invite the guess that x, y and z are all either positive or negative 77 because the second equation would not be so simple otherwise. Make only one of these variable negative and the solution to both equations appears instantly. The problem with these videos is that they focus on a lot of steps rather than on first looking at the likely solution possibilities. This was exactly Feynman's view of algebra--get your answer however it is easiest, not through wrote steps.
y+z=0 or x+y=0 or x+z=0, these are not "and" so we have, if y+z=0 then x=77, from this, we have (1/x)^99+0=(1/77)^99 in this case, we do not have y=-77 and z=77, we have y=-z only
Answer is close to 0
I think this is a wrong approach. If (Y+Z)=0, then (X+Y) doesn't need to be 0. So if X =77, Then Y not necessarily-77.
@infyGyan
Күн бұрын
Yes, from the last three equations x+y=0, y+z=0 & z+x=0 we have to be sure that these won't dis-satisfy the original equation x+y+z=77, so the equation won't satisfy the original one, then we will reject that one. Thanks