Відео

I enjoyed your work and your explanation. Excellent and well done! These problems can be long and it took nearly three pages for this one! Lol Sometimes it feels like you're going down a math rabbit hole with no end but in the end when you do reach a solution (s) it's all good.

Professor, as the X has power of 4, shouldn't the equation have 4 solutions?

I kinda just lucked out, I felt the numbers were too specific so it was probably a round number, I tried subtracting them got nothing, then I tried adding them (1,111,111,088,888,888 + 123,456,787,654,321) and got 1,234,567,876,543,210, which as you can see gives the bottom number multiplied by 10, so if x + y = 10y, then that means x = 9y, the top number is the bottom number multiplied by 9, so the division gives 9, so the answer is 3, the square root of 9

Took me a couple of seconds to determine it is 3... but only because I was looking at repeating number square patterns during the past week. I recognized the top to be the pattern for 3 (1..08..9 the number of 1s or 8 indicates the number of repeating 3s); and the bottom to be the pattern for 1 (123..321 with the highest number at the inflection point indicating the number of repeating 1s)... 3333 3333 / 1111 1111 = 3

328

Nice solution specially how got the (x-3) out of 6x^3+18x^2, neat trick.

X = 3

That would give me writers cramp.....and why would I ever need to do it....

Due to the symmetry of the equations you can deduce that the two x solutions mirror in y: y1 = x2 and y2 = x1.

We know that any odd integer = difference between the square values of two consecutive integers

(1;2) (2;1)

All solutions are 3, 3*e^(2/3pi* i) and 3 *(e^( 4/3pi* i)

Надо сначала вычислить дискриминант, а потом делать вывод о решении квадратного уравнения. Все хорошо, но не рационально решено банальное квадратное уравнение. Зачеркивание в алгебре допускается при уничтожении слагаемых с разными знаками, а вот при сокращении дроби необходимо писать результат, даже если это 1.

👍👍👍😁🤪👋

Phương trình vô nghiệm.

Nice. Anyway it is stupid to use the same symbol (x) for multiplication and as a number.

The technique explained is very lengthy and can be solved easier

Great.

beautiful :)

1.022395

KI (MS-Copilot) solved it in seconds. X=4,723 Y=1,268 or X=1,268 Y=4,723

X= 2.

a= 24, b=23.

X=8 , X=6

Its so easy could be done oraly.

The two-variable equation can be immediately reduced to a one-variable one, due to the symmetries. Let x=3+z, y=3-z. Thus (3+z)(3-z)=36 --> 9-z^2 =36 --> z = sqrt(-27) --> z= i*sqrt(27), ---> x=3+i*sqrt(27), y=3-i*sqrt(27)

Rather easy equations system for Soviet school.

It is immediately apparent that the 4 consecutive multiplicands are around the fourth root of 1680. 1680^(1/4)= 6.4... Let's try: 5*6*7*8 = 1680. (x+2)! falls out -> x+3=5 -> X=2

Author forgot to explain to people familiar with complex numbers that 1 times 1 is 1!

Surprised he didn’t end it with 169 = 13^2

4-

Attendance link... forgot to post it...pl. send it

別在全世界上丟臉了 這種程度,也要貼文,教別人 先去睡個覺吧

You lost the 4rth root? You should have 4 solutions but you missed an opportunity to do another perfect square: (X^2 - (x-4)^2) You squared the right side but could have treated this a perfect square instead. Do you agree?

X=14

X=16

Haven't been doing this for more than thirty years ... and never missed it. 🙂

Nice explain

Very nice. Thank you!

This is the most stupid way of solving this Math Olympiad question.

-7._7=49

Why not just long multiply, add, divide by 2 and then find the square root? Much simpler.

Factorize the cubics in numerator and denominaor and you get the solution almost immediately (without substitutions)

Factorize the sums of cubics in numerator and denominator. Then you get the solution almost immediately.

1

Took be about 10 seconds to divide mentally both num. and den. by 11111111

1.6

16

Help (x+y)^x^1/2=(x-y)^y^1/2

Good