Japanese | Math Olympiad | How to solve for X in this problem ?

Рет қаралды 82,390

4 ай бұрын

Hello My Dear Family😍😍😍
I hope you all are well 🤗🤗🤗
If you like this video about
How to solve this math problem
please Like & Subscribe my channel as it helps me alot ，🙏🙏🙏🙏

Пікірлер: 47

@GaryBricaultLive 4 ай бұрын

By inspection I can see that the numerator and denominator are essentially symmetrical. And the obvious answer being X = 0.

@is7728 4 ай бұрын

Yes I noticed that too

@junkmail4613 4 ай бұрын

I concur

@jim2376 4 ай бұрын

Exactamundo.👍👍

@user-kz3tn5sr6t 3 ай бұрын

@@junkmail4613йффффффффффффффффффффффйф

@prabhudasmandal6429 4 ай бұрын

When we get x^2+71=0 or x^2=--71 ,we can easily say that it is not valid for real value .

@pb2806 4 ай бұрын

Interesting, so the key is to notice that 7+10=17 and 8+9=17, so that we can regroup the terms and obtain similar developments in the form of x^2+17x+constant. Thanks

@user-pd7js7cy9m 4 ай бұрын

If 0

@Nepomenik 4 ай бұрын

Exactly! With added x>0 and not equal to 7, 8 ,9 or 10, this is the most elegant proof.

@scpforjee 3 ай бұрын

5:51 Could have added 1 on both sides, then completed the square. So (a+1)²=(b+1)². Then take square root on both sides. So a+1=±(b+1) and then proceed.

@professorsargeanthikesclim9293 3 ай бұрын

Excellent.

@antonellocossu4319 3 ай бұрын

By looking at the four factors composing the denominator, the solutions can't be 7, 8, 9 and 10. Furthermore, provided there are four factors at the denominator (an even number), and showing the same constants of the factors in the upper side of the fraction, with just their signs changed, it can be deduced that one solution in R is 0, like many comments correctly say below. From observing the equation there's another consequence: the shape of the resulting equation will be x^4+ax^3+bx^2+cx+d = x^4-ax^3+bx^2-cx+d where a, b, c, d >0 and the x terms power 0, 2, and 4 are the same on both sides of the equation. Hence, they will cancel each other, and the resulting equation will have the form ax^3+cx=0 where both a and c will be >0. Therefore, the resulting equation will have the form x(x^2+A)=0 (A=a/c) where A>0, which means that there's one only solution in R: x=0.

@benjaminconcepcion2102 3 ай бұрын

By inspection x=0

@vladsmith3657 4 ай бұрын

Easy, X=0. 👍

@habeebalbarghothy6320 4 ай бұрын

Very clear ! x=0

@prime423 2 ай бұрын

Gary has it right!!Analyze first!!!!!!!

@keinKlarname 4 ай бұрын

x^2 +71 = 0 discriminant to see there are no real solutions?

@samuelbenet007 4 ай бұрын

À cette équation, tu vois tout de suite que S={∅} ^^

@ToanPham-wr7xe 4 ай бұрын

😮

@jim2376 4 ай бұрын

By inspection x = 0 works. (56 x 90)/(56 x 90) = 1. We're positive upstair and downstairs.

@benjaminconcepcion2102 3 ай бұрын

Solvable by the fastest way😭

@herbertklumpp2969 3 ай бұрын

Nothi g to calculate.x=0 is a solution because 7*8*9*10= (-7)*(-8)*(-9)*(-10) For x real x+7 ,> x-7 and Rhein same for the other ( ) therefore nominator is greaterthan denominator conclude nominator/denom. Is 1 für Real x

@olegkalsin5020 4 ай бұрын

Доказать ответ, который лежит на поверхности...

@anjukhandu 4 ай бұрын

I got answer in 1minute by putting value of x ....

@MathSync 3 ай бұрын

I ❤ Mathematics

@learncommunolizer 3 ай бұрын

Thank you very much!

@user-bk1wl9ez1c 4 ай бұрын

Factorial

@chittaranjanmahanta5224 24 күн бұрын

If a/b=1 Then a+1/b+1=1 How it possible?

@comdo777 4 ай бұрын

asnwer=1x

@mauriziocotti9781 3 ай бұрын

Very good

@learncommunolizer 3 ай бұрын

Thank you very much!!

@peterkrauliz5400 3 ай бұрын

There is a second solution if we accept x to be infinitely large.

@user-jl9bs3ip5m 3 ай бұрын

Уважаемые, скажите кому и для каких целей нужны такие уравнения? В школе так вообще замучаешься решать, один пример за урок, на большее времен не хватит.