Since one equation with two unknowns is given, the considered task has infinitely many solutions. Therefore, this task makes sense only if it needs to be solved in integer numbers. So, it is assumed that x and y are integer numbers. Then the approach you have described (i.e. using the notations a=x^(y/2), b=y^(1/2), transforming the original problem into the form (a-b)*(a+b)={1*77; 7*11}) is invalid, since b=y^(1/2) may not be an integer. P.S. Nevertheless, the equation x^y-y=77 can be solved in integer numbers, but not by your false approach, and by a completely different approach! This equation has a finite number of integer solutions! Best wishes, Sharif E. Guseynov Riga, Latvia

THẦY Ở COLOMBIA HAY GERMAN?

(we want to find a p*q=a²+2ab+b; with p and q ∈ N*) So, we will star isolating b and a in the equation to find a divisible term after: For every b and a, not necessarly in Z: a²+2ab+b=44 a²+b(2a+1)=44 2a²+a+b(2a+1)=44+a²+a a(2a+1)+b(2a+1)=44+a²+a (a+b)(2a+1)=44+a²+a a+b= [a²+a+44]/(2a+1) ● b = {[a²+a+44]/(2a+1)}-a; with 2a+1≠0 But, remember b ∈ N*, then: make it: Q(a)=a²+a+44 and q(a)=2a+1 then, [a²+a+44]/(2a+1)=[Q(a)]/[q(a)]=p(a) and p(a)=n, n ∈ N*. /////////////////////////////////////////////////// finding the p(a) and r(a): Q(a) = a² + a + 44 | q(a) =(2a+1) -a² - a/2 ||p(a)=(a/2)+(1/4) -----------‐ 0 + a/2 + 44 - a/2 - 1/4 ------------------- 0 + 175/4 = r(a) ///////////////////////////////////////////////// Constructing p*q into the original equation: So, Q(a)=q(a)*p(a) + r(a) ● a²+a+44=(2a+1)*[(a/2)+(1/4)] +(175/4) retake the original equation: a²+2ab+b=44 a²+b(2a+1)=44 2a²+a+b(2a+1)=44+a²+a a(2a+1)+b(2a+1)=44+a²+a (a+b)(2a+1)=44+a²+a; but remember Q(a)=44+a²+a then, (a+b)(2a+1)=(2a+1)[(a/2)+(1/4)]+(175/4) make 4 times the equation: (4a+4b)(2a+1)=(2a+1)[2a+1]+175 (4a+4b)(2a+1)=(2a+1)²+175 (4a+b)(2a+1) - (2a+1)² =175 (2a+1)[(4a+4b)-(2a+1)]=175 (2a+1)[4a-2a+4b-1]=175 (2a+1)(2a+4b+1)=175 Where are the p and q? It is here the term p=(2a+1) and it the term q=[(2a+1)+4b]. ■(2a+1)[(2a+1)+4b]=175; n=2a+1 better make like this way: ■n(n+4b)=175 Now, we can thing the possibilites: n×(n+4b)=5²×7; with ■D(175)={1;5;7;25;35;175} And ■ n ⊂ D(175) (I) then, n ⊂ {1;5;7;25;35;175} ////////////////////////////////////////////////// Considering the domain of the question: remember that n=2a+1 and consider a ∈ N* because the question already said it. And, remember b ∈ N. And think about a domain of a and b. So, retake a²+2ab+b=44 you can see that each term of the equation cannot be bigger than 44. Then, a²<44; 2ab<44; b<44 taking a²<44 and a ∈ N* then, 0<a<7 but a ∈ N* Therefore, ■ 6≥a≥1 retake n=2a+1 and take 0<a<7. then, ■13≥n≥3. (II) /////////////////////////////////////////////// take (I) and (II): n=5 or n=7 therefore, a=2 or a=3 n×(n+4b)=[D(175)]×[D'(175)] input a={2;3} into equation: a²+2ab+b=44 a=2 2²+2*2*b+b=44 4+4b+b=44 5b=41 b=41/5; but b belongs to N* so, a=2 is not a solution. a=3 3²+2*3*b+b=44 9+6b+b=44 7b=35 b=5 Therefore, ■ a=3 and b=5 is the solution. ------------‐------- / / -----‐------------

a²+2ab+b=44 a, b ∈ N Then, a² ∈ N, ab ∈ N, (a-b) ∈ Z and (a+b) ∈ N. then, a²<44, 2ab<44, b<44 so, ■a²<44 => a<√44<7 => a<7 then 0<a<7 and a ∈ N ● 6 ≥ a ≥ 1with a ∈ N ■2ab<44 => ab<22 ■b=44-a²-2ab if a=1 (the smalest a possible) then, b=44-1-2b 3b=43 b=43/3≈ 7 (the biggest b); but b ∈ N, then 7≥ b>0 if a=6 (biggest) then, b=44-36-6b 7b=8 b≈1 (smalest) then, ● 7 ≥ b ≥ 1 with b ∈ N ■ab ---> 7*1=7. (smallest ab) 6*2=12 5*3=15 4*4=16 (biggest ab) then, 14 < 2ab < 32 So, a²+2ab+b=44 a²+b=44-2ab 44-14 > a²+b > 44-32 30 > a²+b > 12 then, 30 - b > a² > 12 - b 30 -1 > a² > 12 - 7 29 > a² > 5 36 > 29 > a² > 5. > 4 6² > a² > 2² 6 > a > 2 with a ∈ N then, 5 ≥ a. ≥ 3 input a={3;4;5} in the equation: a²+2ab+b=44 when, a=3 3²+2*3b+b=44 9+6b+b=44 7b=35 => b=5 so, a=3 and b=5 is a solution a=4 4²+2*4b+b=44 16+8b+b=44 16+9b=44 9b=28 b≈7 is not a solution a=5 5²+2*5b+b=44 25+10b+b=44 25+11b=44 11b=19 is not a solution. Therefore, a=3 and b=5 is the anwser.

Once getting (m+1)^4 - (m-1)^4 = 80，unfolding left side and merging them. 2* (4m^3 + 4m) = 80，then m^3 + m = 10. Guessing m = 2 is one of solutions....

WHAT HE FUCK THIS NIS NOT MATH OLYMPIAD LEVEL WHAT IN THE ACTUAL FUCK ARE YOU ON ABOUT THIS IS SO FUCKING DUMB, STOP MAKING HTESE SHITTY CLICKBAITS

Se tivesse feito 6^x = 3^x *2^x , e a = 3^x; b=2^x, ficaria bem mais simples. Quadrar e dividir por ab. Chegaria em a/b + b/a - 3 = 0, Fazendo a/b = k e ... ( If I had done 6^x = 3^x *2^x , and a= 3^x; b =2^x, it would be much simpler. Square and divide by ab. It would arrive at a/b + b/a - 3 = 0, making a/b = k and...).

Ngủ sớm. Mọi ý tưởng cạn kiệt

Solution: 4^x-8*x = 0 = 4^2-8*2 |The same operations are done on the left side of the equation with x as on the right side of the equation with 2, therefore x = 2. When you use the complicated Lambert function, you make also a comparisation between the left side of the equation with the right side of the equation. And you can save yourself this complicated comparison if you make this comparison right away.

Đặt U = . V= -. (U+V)^3 =? (U = a^1/2 .V =b^1/2)

KHÔNG ĐỦ KIÊN NHẪN. NHỜ THẦY GIÚP

XEM VIET CÓ OK KHÔNG THẦY.

(27^x)+x=0 --> 27^x=-x Raise to 1/x and noting that 27=3³: 3³=(-x)^(1/x) =(1/t)^(-t) by letting t=-1/x =[(1/t)^(-1)]^t =t^t --> t=3 -1/x=3 --> x=-⅓

Hằng Đẳng thức hả thầy.

Решил устно. Конечно,знал метод.

x=3

(x-1)½+(x+2)½=3; make it: u=(x-1)½ and v=(x+2)½ then, u+v=3 (u+v)(u-v)=3(u-v) u²-v²=3(u-v); but u²=(x-1) and v²=x+2 then, (x-1)-(x+2)= 3(u-v) -3=3(u-v) u-v=-1 make the sistem: u+v=3 u-v=-1 then, 2u=2 therefore, u=1 and v=2 But u=(x-1)½ and v=(x+2)½ then, (x-1)½=1 and (x+2)½=2 therefore, x-1=1 => x=2 and x+2=4 =>x=2 therefore, x=2 is a solution. -------‐-------‐------- / / -------‐--------‐‐‐-- simplify one term way: (x-1)½+(x+2)½=3; u=x+2 =>x-1=u-3 (u-3)½+u½=3 (u-3)½=3-u½ (u-3)½=(-1)[(u½)-3] [(u-3)½]²={(-1)[(u½)-3]}² u-3=(-1)²[(u½)-3]² u-3=[(u½)-3]² u-3=(u½)²-2*(u½)*3+3² u-3=u-6(u½)+9 6(u½)=12 u½=2 u=4 but u=x+2; then 4=x+2. therefore x=2 ---------------------- / / --------------------- cheating way: (knowing √1=1) (x-1)½+(x+2)½=3 when x-1=1, then (x-1)½=1 because √1=1. So, 1+√(2+2)=3 1+√4=3 1+2=3 true. therefore x=2 is a solution. ----‐-------------- / / --------------------- algebrical way: (x-1)½+(x+2)½=3; make it: a=(x-1)½ and b=(x+2)½ then a+b=3 b = 3-a b(3+a) = (3-a)(3+a) 3b+ab=3²-a²; but 3=a+b (a+b)b+ab=9-a² ab+b²+ab=9-a² 2ab=9-(a²+b²) 2√(x-1)√(x+2)=9-[(x-1)+(x+2)] 2√[(x-1)(x+2)]=9-[2x+1] 2√[x²+(-1+2)x+(-1)(2)]=9-2x-1 2√[x²+x-2]=8-2x 2(x²+x-2)½=2(4-x) (x²+x-2)½=(4-x); make it: c=(x²+x-2)½ and b=(4-x) c=d c*d=d*d cd=d², but c=d; then cd=c² therefore, c²=d² [(x²+x-2)½]²=(4-x)² x²+x-2=(4²-2*4*x+x²) x²+x-2=(16-8x+x²) x²+(x-2)=x²+(16-8x) (x-2)=(16-8x) (x-2)=8(2-x) (x-2)=8(-1)(x-2) (1)(x-2)=(-8)(x-2); z=x-2 [1)*z=(-8)*z is true only if: z=0 then, x-2=0 x=2 is a solution.

√[6 - √(6 + x)] = x x ≥ 0 √(6 + x) = u √(6 - u) = x 6 + x = u² 6 - u = x² u² - x² = u + x (u + x)(u - x) - (u + x) = 0 (u + x)(u - x - 1) = 0 u = - x ∨ u = x + 1 u = -x => 6 + x = x² x² - x - 6 = 0 (x - 3)(x + 2) = 0 x = 3 ∨ x = -2 [ both are invalid ] u = x + 1 => 6 - (x + 1) = x² x² + x - 5 = 0 *x = (-1 + √21)/2*

ab equals 40, so m×m+1=a, and 2m=b

You videos are Brilliant!!! you explain these solutions so well thank you 🙂

Answer=25

4ab=(a+b)^2-(a-b)^2 formula should have been applied without going into so much complexity

You forgot to explain the reasoning behind the various steps, in particular the first one.

3:57 "it is obvious that y=3" NOOO! It is obvious that y=3 IS one of the solutions, but why it is the one and only? It is not obvious! I can say more, if you replace 27 with 0.99 there will be 2 solutions. You HAVE TO JUSTIFY that there is only one solution in this case.

You make things very lengthy at time. When you found (m^2+1)(2m) =20. Its clear that (2^2+1).2.2 = 5.4 = 20. Thus m = 2.

case 1 , -ln2/4= -2*ln2*x*e^(-2*ln2*x) , /4/4 * -ln2/4 = 4*-ln2/16 , --> (4*-ln2)*e^((-ln2)^4) , (4*-ln2)*e^(4*-ln2) , / 4*-ln2=2*-ln2*x , -4=-2x , X1=2 , test x1 , OK , case 2 , W(-ln2/4)=-2*ln2*x , -0.2148111164=-2*ln2*x , x2=-0.2148111164/(2*ln2) , test x2 , OK ,

A questão é muito boa, mas eu demorei uns 10 minutos para encontrar uma solução simples e muito diferente da sua. Parabéns pela escolha. Brasil - Outubro de 2024. The question is very good, but it took me about 10 minutes to find a simple solution that is very different from yours. Congratulations on your choice. Brazil - October 2024.

I like it very much of solution

You have to prove the hypothesis. a=7x²,b=7y²

If the original equation is transformed into an equation that has the one variable on one side of the equation and the other variable on the opposite side of the equation, we will have: b=(44-a^2)/(2a+1). From the above relationship, if we assign any value for "a", we will obtain a corresponding value for "b". Therefore, there are infinitely many pair of solutions for "a" and "b". Examples are: (a,b)=(0,44), (2,8), (3,5), (12,-4), (17,-7), (87,-43), (-1,-43),(-3,-7), (-4,-4), (-13,5), (-18,8), (-88,44) and so on.

2

log(5)x * log(5)x = log(5)625 , 5^4=625 , log(5)625=4 , (log(5)x)^2=2^2 , log(5)x=+/-2 , x= 5^2 , 5^(-2) , x= 25 , 1/25 ,

Если знать таблицу квадратных чисел, то такую задачу можно решить за несколько сикунд 25^2=625

x=(-3) ?

The method used is rather famously called completing the product or SFFT(Simons favorite factoring trick) Great way create common binomial factors. thanks

x = 2

x = (25 , 1/25)

x^[log_5(x)]=625 [5^[log_5(x)]]^[log_5(x)]=625 Let n=log_5(x) (5ⁿ)ⁿ=625 5^(n²)=5⁴ n²=4 |n|=2 n=-2 log_5(x)=-2 x=5^-2 x=0.04 ❤ n=2 log_5(x)=2 x=25 ❤

W(-(ln4)/8)=W(-ln8xe^(-ln8x)) gives both solution simultaneously

X=25

M=-1/4

Note that there are two unknowns in the equation. With no constraint being imposed there are many solution. If (x,y) is a pair of integer the trivial solutions: (x^y)-y=77 =78¹-1 --> (x,y) --> (78,1) =1^(-76)-(-76) --> (1,-76) Other solution: (x^y)-y=81-4 (x^y)-y=3⁴-4 --> (x,y)=(3,4) by comparing the structure. The general approach is [x^(½y)]²-(y^½)²=77 [x^(½y)+(y^½)][x^(½y)-(y^½)]=7×11 x^(½y)+y^½=11 x^(½y)-y^½=7 Half of the sum is x^(½y)=9 Half of the difference is y^½=2 --> y=4. Hence x=3 --> (x,y)=(3,4 ) Thus (x,y)={(-1,76),(3,4),(78,1)}

Que que você fez uma questão bonita. Parabéns 👏👏👏👏👏

X=-1/3

You solve in a unique way different from other pls can you solve the equation below a + b + c = 6 a² + b² + c² = 14 a⁸ + b⁸ + c⁸ = ?

First method >>>>>>

Good effort

Excelente solución

Nice solution sir

There are infinite solutions. (x, y) = (78, 1), (√79, 2), (80^(1/3), 3), (3, 4), (82^(1/5), 5), (83^(1/6), 6), ・・・　[ (-3, 4) is also one of the solutions. ]