You missed the solutions should give the set of number: The set of numbers should be specified. With (x,y) ∈ ℤxℤ You get: (x , y) ∈ {(-8,-5), (-8,5),(8,-5),(8,5), (-20,-19),(-20,+19),(20,-19),(20,19)} With (x,y) ∈ QxQ, you will get an infinite number of solutions.

Ill-posed problem. General (infinite) solution in the reals: all the points (x,y = ±√(x^2 - 39)) with |x|≥√39 Infinite solutions can also be given in complex numbers

You say solve the "Math Olympiad problem" but then magically make it so that x-y<x+y. So, why not at the beginning of the problem state that we are looking for specific values of x,y? We are just suppose to throw out solutions like (-20,-19), (-8,-5), (-20,19), (20,-19), etc...?

m^2*m=-8 m=-2 m=1±Sqrt[3]i

Yes , I solved it in my mind within 30 seconds 8^2 - 5^2 64 - 25 = 39

I learned this method from you, thank you! , x^2-y^2=39*1 , (x+y)(x-y)=39*1 , let x+y=39 , let x-y=1 , x+y+x-y=39+1 , 2x=40 , x=20 , y=x-1 , y=20-1 , y=19 , test , 20^2-19^2=400-361 , 400-361=39 , OK , --> , 39= 3*13 , 13*3 , 1*39 , x^2-y^2=13*3 , (x+y)(x-y)=13*3 , let x+y=13 , let x-y=3 , x+y+x-y=13+3 , 2x=16 , x=8 , y=x-3 , y=8-3 , y=5 , test , 8^2-5^2=64-25 , 64-25=39 , OK , x^2-y^2=3*13 , (x+y)(x-y)=3*13 , let x+y=3 , let x-y=13 , x+y+x-y=3+13 , 2x=16 , x=8 , y=x-13 , y=8-13 , y= -5 , test , 8^2-(-5)^2=64-25 , 64-25=39 , OK , x^2-y^2=1*39 , (x+y)(x-y)=1*39 , let x+y=1 , let x-y=39 , x+y+x-y=1+39 , 2x=40 , x=20 , y=1-20 , y= -19 , test , 20^2-(-19)^2=400-361 , 400-361=39 , OK ,

Why is middle school-grade problem is called olympiad?

x = 3 y=6 is also a solution, you can reach it by 8 = (1/2) ^ (-3)

Não é possível ver a resposta final, o vídeo termina antes de terminar.

wonderful sir great explanation thank you very much!

By writing -810 as -729-81, I could factor the equation by grouping, which is the distributive property in reverse.

And then there's me, who got y = 3 by just testing y=1 and up

let u=2^x , u^3 + u^2 + u - 39=0 , (u-3)(u^2+4u+13)=0 , u=3 , u^2+4u+13=0 , +1 -3 u=(-4+/-V(16-52))/2 , u=(-4+/-V(-36))/2 , u=(-4+/-i*V(36))/2 , +4 -12 u= 3 , / -2+3i , -2-3i /, 2^x=3 , x=log3/log2 , test , x=log3/log2 , +13 -39=0 , 2^(log3/log2)+4^(log3/log2)+8^(log3/log2)=3+9+27 , --> 39 , OK

What was all that 6 minutes of video for? You just went guessing in the end, totally forgetting about negative or fraction possibilities.

65520. 4^8 is (4^2)^2)^2 or (16^2)^2 or 256^2 = 65536. 65536 - 16 = 65520.

at time 0:54 - 2^16 - 16. Since 2^16 is, as we were all taught in computer kindergarten, 65, 536, we can just subtract 16 to get 65, 520.

Get yourself a calculator 😭😂

lucky it was 2 otherwise forget it, poor method

16,366 Mental calculation

Is guessing acceptable in the exam? I don't know...

wow. I was not familiar with your working but it agrees with open ai's opinion. open ai says in complex numbers, we can define the logarithm of a negative number. It says -2 = 2 . -1 = 2 . e^( i pi) log(-2) = log(2) + i pi. In fact log(-2) has multiple values as you can add 2 n pi i where n is any integer

Working in unreal spaces is cool, and I wish there were more graphic visualizations of problems like this. Do you know of a way generating them?

It’s in my head.

5^m=m m=-(W_n(-ln(5)))/ln(5), n element Z

There is no solution. 1 to whatsoever is 1

Bruh in which country is this an olympiad problem? 😂

1^x=-2 x=-(Ln(Surd[2,2πK])+(0.5+0.5π)i)/K)=-(Ln(Surd[2,2πK])+(1+π)i/(2K)) K=Z Z is any integer final answer

xy=32=1×32=2×16=4×8 Here 4+8=12 (x,y)=(4,8) or (8,4)

x = 8 y = 4 8 + 4 = 12 8 x 4 = 32

we get , m^3=64 , m^3 +/- m^2 +/- m - 64=0 , m^2(m-4)+4m(m-4)+16(m-4)=0 , (m-4)(m^2+4m+16)=0 , m=4 , +1 -4 m^2+4m+16=0 , m=(-4 +/- V(16-4*16))/2 , m=(-4 +/- V(-48))/2 , +4 -16 m= -2 + i*V12 , -2 + i*V12 , +16 -64=0 , test with m=4 , 4/(m) * 4/(m) = (m)/4 , 4/(4) * 4/(4) =1*1 , (4)/4=1 , OK ,

x=32/y , --> , y^2 - 12y + 32=0 , --> , y^2-8y -4y+32=0 , (y-8)(y-4)=0 , y= 8 , 4 , x=12-y , case 1 , y=8 , x=12-8 , x=4 , +1 -8 case 2 , y=4 , x=12-4 , x=8 , -4 +32=0 , solu , (x , y) , (4 , 8) , (8 , 4) , test , x*y=32 , 4*8=8*4 , --> , 32 , OK ,

5^x. 5^x. 5^x = 10 X log 5 + X log 5 + X log5 = 10 3X log 5 = 10 3X = 10 ➗ log 5 3X = 10 ➗ .70 X = (1000/70)➗ 3 = ~5

c

It's 12:01am...

tan(x) = AC/AB tan(x) = (3 + a)/b Triangle ABD → AB² + AD² = BD² b² + a² = 5² a² + b² = 25 Triangle ABC → AB² + AC² = BC² b² + (3 + a)² = 7² b² + 9 + 6a + a² = 49 a² + b² + 6a = 40 → recall: a² + b² = 25 25 + 6a = 40 6a = 40 - 25 6a = 15 a = 15/6 a = 5/2 Recall: a² + b² = 25 b² = 25 - a² b² = 25 - (5/2)² b² = (100/4) - (25/4) b² = 75/4 b = ± √(75/4) ← as b represents a length, we keep only the positive value b = √(75/4) b = √[3 * (25/4)] b = √[3 * (5/2)²] b = (5/2).√3 b = (5√3)/2 Recall: tan(x) = (3 + a)/b → where: a = 5/2 tan(x) = [3 + (5/2)]/b tan(x) = [(6/2) + (5/2)]/b tan(x) = (11/2)/b → where: b = (5√3)/2 tan(x) = (11/2)/[(5√3)/2] tan(x) = (11/2) * [2/(5√3)] tan(x) = 11/(5√3) tan(x) = (11 * √3)/(5√3 * √3) tan(x) = (11√3)/15 → x = 51.7868 °

Nailed it. Thanks for the challenge.

x=(1+Log[5,2])/3 final answer

5^x*5^x*5^x=10 x=Log[125,2]+0.3 recurring=Log[125,2]+1/3

*m^3 = 4^3* *(m-4)(m^2+4m+16)=0* *m=4 hoặc m=-2 ± 2i√3*

😅😅😅😅 good

Cool😊

4/4*4/4=4/4 x=4 x=-2+2Sqrt[3]i x=-2-2Sqrt[3]i

Easy

??????😂

You added 1 chocolate😂

m^2-n^2=9 , (m+n)(m-n)=9*1 , let m+n=9 , & m-n=1 , m+n+m-n=9+1 , 2m=10 , m=5 , n=m-1 , n=5-1 , n=4 , test , 5^2-4^2=25-16 , 25-16=9 , same , OK ,

6x^3-48=0 x=2 x=-1±Sqrt[3]i

m^3=1 m=1 m=-0.5 ±0.5Sqrt[3]i=(-1±Sqrt[3]i)/2 It’s in my head.

m^3 +/- m^2 +/- m -1=0 , +1 -1 m^2(m-1)+m(m-1)+(m-1)=0 , (m-1)(m^2+m+1)=0 , m=1 , +1 -1 m^2+m+1=0 , m=(-1+iV3)/2 , (-1-iV3)/2 , +1 - 1=0 , solu. , m= 1 , (-1+iV3)/2 , (-1-iV3)/2 ,

Cool number designs for a rat.