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X=(4.39) approximately ....May be Explain later

I think those answer is not wrong, but not true cause root. Exactly plus minus if the rule of root.

Thanks for sharing all these math challenges. I wish you & the followers all the best for 2025!

All substitutions are like You had an epiphany. Ridiculous problem

I see your video runs almost 18 minutes, but I noticed x = 2 in seconds. It is important to know Lambert-W, but also important to be able to notice things. Nobody can spare 18 minutes for an admission question.

thank you very much，sir

√m + √n = 9 ← this is the sum S √m * √n = 9 ← this is the product P √m & √n are the solution of the following equation: x² - Sx + P = 0 → where S is the sum and where P is the product x² - 9x + 9 = 0 Δ = (- 9)² - (4 * 9) = 81 - 36 = 45 x = (9 ± √45)/2 x = (9 ± 3√5)/2 ← these are √m & √n First case: √m = (9 + 3√5)/2 (√m)² = [(9 + 3√5)/2]² m = (9 + 3√5)²/4 m = (81 + 54√5 + 45)/4 m = (126 + 54√5)/4 m = (63 + 27√5)/2 m = 9.(7 + 3√5)/2 → then you can deduce that: n = 9.(7 - 3√5)/2 Second case: √m = (9 - 3√5)/2 (√m)² = [(9 - 3√5)/2]² m = (9 - 3√5)²/4 m = (81 - 54√5 + 45)/4 m = (126 - 54√5)/4 m = (63 - 27√5)/2 m = 9.(7 - 3√5)/2 → then you can deduce that: n = 9.(7 + 3√5)/2

x = approx. -0.371

Why do you always make maths more complicated than it is. You should just solve X^2-y^2=sqrt(10) with X^2+y^2=sqrt(50) simultaneously. And get their square root after. Very silly

√[3^(x)].√[3^(x)].√[3^(x)].√[3^(x)] = 2¹⁵ → Let: a = 3^(x) √a.√a.√a.√a = 2¹⁵ → you know that: √a = a^(1/2) √a.√a.√a.[a^(1/2)] = 2¹⁵ → you know that: a.[a^(1/2)] = a^(1) * a^(1/2) = a^[1 + (1/2)] = a^(3/2) √a.√a.√a^(3/2) = 2¹⁵ → recall: √a = a^(1/2) √a.√a.[a^(3/2)]^(1/2) = 2¹⁵ → you know that: [x^(a)]^(b) = x^(ab) √a.√a.[a^(3/4)] = 2¹⁵ → you know that: a.[a^(3/4)] = a^(1) * a^(3/4) = a^[1 + (3/4)] = a^(7/4) √a.√[a^(7/4)] = 2¹⁵ √a.[a^(7/4)]^(1/2) = 2¹⁵ √a.[a^(7/8)] = 2¹⁵ → you know that: a.[a^(7/8)] = a^(1) * a^(7/8) = a^[1 + (7/8)] = a^(15/8) √a^(15/8) = 2¹⁵ [a^(15/8)}^(1/2) = 2¹⁵ a^(15/16) = 2¹⁵ → recall: a = 3^(x) [3^(x)]^(15/16) = 2¹⁵ [3^(x)]^(1/16) = 2 3^(x) = 2¹⁶ Ln[3^(x)] = Ln(2¹⁶) x.Ln(3) = 16.Ln(2) x = 16.Ln(2) / Ln(3)

x = 16 log 2/log 3

1/2 min vs 15.3 min. but money

3^((15/16)*x)=2^15 , 3^((15/16)*x)=3^((ln2/ln3)*15) , (15/16)*x=(ln2/ln3)*15 , x=(ln2/ln3)*15/(15/16) , x=(ln2/ln3)*16 , test , 3^((15/16)*x)=~ 32768 , 2^15=32768 , same , OK ,

X=16(log2/log3).....May be Explain later....... As per question 3^{(x/2)+(x/4)+(x/8)+(x/16)=2^15 3^(15x/16)=2^15 {3^(x/16}^15=2^15 So, 3^(x/16)=2 Take the log log{3^(x/16)}=log2 (X/16).log3=log2 (X/16)=log2/log3 X=16{log2/log3}

Rule no.1 put x=0. Rule no.2 put x=digits in the equation. Rule no.3 use your brain Rule no.4 Don't think all questions tagged Olympiad are not Olympiad level. I got solution from rule 1 and 2😂

Посчитать кубический корень из 4913 без калькулятора. Нет ничего проще, конечно же - 17. Удивительно, почему он сразу ответ без калькулятора не написал.

Roughly 90% of the "simplification" was not necessary, sorry.

Your solution is more rigorous but here's a quick alternative. Also I would have wanted to use a calculator to evaluate the arithmetic that led to xy = 17 and it was not allowed. It's unlikely that the √2 will not be in the solution. If there is a solution, it's probably going to be of the form a + b√2. So (a + b√2)^3 = 55 + 63√2 Expanding a^3 + 3a^2*b√2 + 3ab^2*2 + b^3*2√2 = 55 + 63√2 For integer a & b, this means that a^3 + 6ab^2 = 55 Factoring a(a^2 + 6b^2) = 55 The terms inside the parentheses will be positive, so a must be >0. Also a must be < 55^(1/3), which means that a = 1, 2, or 3. The a, (a^2 + 6b^2) possibilities are 1,55 or 5,11 etc. The only one that works is a=1. Then b=3. The other half of the first expansion is 2b^3√2 + 3a^2*b√2 = 63√2 Eliminating the √2 gives 2b^3 + 3a^2*b = 63 Or b*(2b^2 + 3a^2) = 63 The a=1 and b=3 that were previously found satisfy this equation. Thus the solution is 1 + 3√2

Why can't we use common sense and solve such problems in a flash? Something power four is about 240 means that is 4. So 7+1/x=4 and so x= -1/3.

At about 12:00 "Case 2" x² + y ² = -√(50). Surely the LHS is positive and this must be rejected, unless you are about to entertain complex solutions. Expressions like √(2 + √5) can sometimes be simplified as "a + b√5", altho I have forgotten what the limitations are. Please make a video on denesting radical expressions.

X,2×+5=8[n3]

[7+(1/x)]⁴-[5+(1/x)]⁴=240 Let y=(1/x) (y+7)⁴-(y+5)⁴=240 (y+6+1)⁴-(y+6-1)⁴=240 Let z=y+6 (z+1)⁴-(z-1)⁴=240 z⁴+4z³+6z²+4z+1 -z⁴+4z³-6z²+4z-1 8z³+8z=240 8(z³+z)=240 z³+z=30 z³+z-30=0 z³-27+z-3=0 (z-3)(z²+3z+9)+1(z-3)=0 (z-3)(z²+3z+10)=0 z²+3z+10=0 4z²+12z+40=0 4z²+12z+9=-31 (2z+3)²=-31 |2z+3|=i√31 2z+3=±i√31 2(y+6)+3=±i√31 2y+12+3=±i√31 2y+15=±i√31 2y=-15±i√31 (2/x)=-25±i√31 x(-25±i√31)=2 x=[2/(-25±i√31)] ❤❤ z-3=0 z=3 y+6=3 y=-3 (1/x)=-3 -3x=1 x=-⅓ ❤

My approach: a=7+1/x; b =5+1/x ---> a-b=2 and a^4 - b^4 =240 --->(a^2-b^2)(a^2+b^2)=240 -----> 2(a+b)(4+2ab)=240 ----->(a+b)(2+ab)=60 ---->a^3-3a^2+4a-32 =0 a=4 is a solution ----> a=4; b=2 ---->x=-1/3

Too easy for Harvard

Can you please explain how did you anticipate "y" value In the fist step?

16^2 - 4^2 =240 [(4^2]^2 - [(2^2]^2 = 240 4^4 - 2^4 = 240 (7 + 1/x)^4 - (5 + 1/x)^4 = 240 7 + 1/x = 4 .(x) 7x+1=4x x= 1/3 5 + 1/x = 2 .(x) 5x+1=-2x x= -1/3 S = { -1/3}

Porque copias otra vez el enunciado?

Ridiculous complication.

let u=5+1/x , we get , u^3+3u^2+4u-28=0 , (u-2)(u^2+5u+14)=0 , u=2 , 2=5+1/x , x= -1/3 , 1 -2 test , (7+1/(-1/3))^4-(5+1/(-1/3))^4=(7-3)^4-(5-3)^4 , 4^4-2^4=256-16 , --> 240 , OK , 5 -10 solu , x=-1/3 , 14 -28

Thank you for breaking this down so well! You're an inspiration to get back into studying the subject!❤

If you were to multiply by conjugate, why go through the simplifying of √108?

Take 6+1/ x =z and expand( z+1)^4-(z-1)^4 which would give z^3+z=30. Rest is easy.

x² - y² = √10 → given: xy = √10 → x²y² = 10 → y² = 10/x² x² - (10/x²) = √10 (x⁴ - 10)/x² = √10 x⁴ - 10 = x²√10 x⁴ - x²√10 - 10 = 0 Δ = (- √10)² - (4 * - 10) = 10 + 40 = 50 = 2 * 5² x² = (√10 ± 5√2)/2 First case: x² = (√10 + 5√2)/2 Recall: y² = 10/x² y² = 10/[(√10 + 5√2)/2] y² = 20/(√10 + 5√2) y² = 20.(√10 - 5√2)/[(√10 + 5√2).(√10 - 5√2)] y² = 20.(√10 - 5√2)/[10 - 50] y² = 20.(√10 - 5√2)/- 40 y² = (5√2 - √10)/2 x² + y² = [(√10 + 5√2)/2] + [(5√2 - √10)/2] x² + y² = [(√10 + 5√2) + (5√2 - √10)]/2 x² + y² = (√10 + 5√2 + 5√2 - √10)/2 x² + y² = (10√2)/2 x² + y² = 5√2 Restart x² - y² = √10 (x + y).(x - y) = √10 (x + y)².(x - y)² = 10 (x + y)².(x² + y² - 2xy) = 10 → recall: x² + y² = 5√2 (x + y)².(5√2 - 2xy) = 10 → given: xy = √10 (x + y)².(5√2 - 2√10) = 10 (x + y)² = 10/(5√2 - 2√10) (x + y)² = 10.(5√2 + 2√10)/[(5√2 - 2√10).(5√2 + 2√10)] (x + y)² = 10.(5√2 + 2√10)/[50 - 40] (x + y)² = 5√2 + 2√10 x + y = ± √(5√2 + 2√10) Second case: x² = (√10 - 5√2)/2 ← this is a negative number → x is a complex number Recall: y² = 10/x² y² = 10/[(√10 - 5√2)/2] y² = 20/(√10 - 5√2) y² = 20.(√10 + 5√2)/[(√10 - 5√2).(√10 + 5√2)] y² = 20.(√10 + 5√2)/[10 - 50] y² = 20.(√10 + 5√2)/- 40 y² = - (√10 + 5√2)/2 ← this is a negative number → y is a complex number x² + y² = [(√10 - 5√2)/2] + [- (√10 + 5√2)/2] x² + y² = [(√10 - 5√2) - (√10 + 5√2)]/2 x² + y² = (√10 - 5√2 - √10 - 5√2)/2 x² + y² = - (10√2)/2 x² + y² = - 5√2 Restart x² - y² = √10 (x + y).(x - y) = √10 (x + y)².(x - y)² = 10 (x + y)².(x² + y² - 2xy) = 10 → recall: x² + y² = - 5√2 (x + y)².(- 5√2 - 2xy) = 10 → given: xy = √10 (x + y)².(- 5√2 - 2√10) = 10 (x + y)² = - 10/(5√2 + 2√10) (x + y)² = - 10.(5√2 - 2√10)/[(5√2 + 2√10).(5√2 - 2√10)] (x + y)² = - 10.(5√2 - 2√10)/[50 - 40] (x + y)² = (5√2 - 2√10).i² x + y = ± i√(5√2 - 2√10) Summarize: x + y = ± √(5√2 + 2√10) x + y = ± i√(5√2 - 2√10)

RHS to be written as 256-16

Idk whether I'm wrong but can we do it this way? (1/8)^1/8= (2^-3)^1/8 =[(2^2)/(2^5)]^1/8 =[(2^1)/(2^5)]^1/4 = (2^-4)^1/4 =2^-1 = 1/2

^=read as to the power *=read as square root Now explain 9× *{3.(*3) =3^2 ×{3^(1/2)}×{3^(1/4)} =3^(11/4) As per question {3^(11/4)}^(1/5^x) =3^(11/4.5^x) Now 243=3^5 As per question 3^{11/4.5^x}=3^5 So, 11/4.5^x=5 So, 11/4=5^x. 5=5^(x+1) So, 5^(x+1)=11/4 Take log log{5^(x+1)}==log (11/4) (X+1).log5=log11-log4 X+1={log11-log4}/log5 X={(log11-log4 )/log5}-1..May be

(3^(11/8))^(1/(5^x))=3^5 , (11/8)^(1/(5^x))=5 , / ()^(5^x) , 11/8=5*5^x , 5^x=11/40 , x=log(11/40)/log(5) , test , (3^(11/8))^(1/(11/40))=3^(11/8 * 40/11) , --> 3^5=LHS , RHS --> 243=3^5 , same , OK ,

Easy ❤❤

For taking the log to base 5 as it was the appropriate log base in this case, it is better to cancel the log of base and the exponential to reduce the number of steps to solve exponentials when the log base is appropriate. The power rule is implicit when doing such a move.

Отличное решение

(x-5)(x-3)=[24/(x(x+2))] Domain x(x+2)≠0 x≠{-2, 0} x(x+2)(x-5)(x-3)=24 (x²-3x)(x²-3x-10)=24 Let y=x²-3x y(y-10)=24 y²-10y=24 y²-10y-24=0 (y+2)(y-12)=0 y-12=0 x²-3x-12=0 4x²-12x-48=0 4x²-12x+9=57 (2x-3)²=57 |2x-3|=√57 2x-3=±√57 2x=3±√57 x=½(3±√57) ❤❤ y+2=0 x²-3x+2=0 (x-1)(x-2)=0 x-1=0 x=1 ❤ x-2=0 x=2 ❤

There is a second solution for roots of y = 3^(x^2) - x^9. Newton-Raphson (whatever method you memorized) gets you to approx 1.188 pretty quickly. I started at 1 because it is easy to see that the function is positive at x=1, but negative at x=2. (So there ARE starting points that just take you back to 3, and you want to avoid them.) It would be interesting to know what Wolfram says.

Substitute x=y+3/2 into the given equation and rearrange to (y^2-9/4)(y^2-49/4)-24=y^4-(29/2)y^2+441/16-384/16=y^4-(29/2)y^2+57/16=0. y^2=(29/2±28/2)/2=57/4 or 1/4; y=±√57/2 or ±1/2 and x=3/2+y=(3±√57)/2 or 2 or 1.

Are you kidding me? I would solve much faster than that without using the calculator. 3125*3125*5-9

SI SALVE SONO VLADIMIR POPOVIČ HO SPIEGATO A SATANA ((11111×(999+3)^(1/3)×3)÷10))=33335,207201772 √1111222225=33335 33335×33335=1111222225 33335²=1111222225 1,11122225×10⁹=1111222225 √a=b² QUESTO È UN BEL VIDEO.

The solution is best described as pathetic. No need to substitute with x and y. Do without the substitution. Also, the smart people reading this mistakenly devised solution don't need to be misled. And that's all this writeup does.

Your work was clumsy! There is no need to expand the calculation to so many steps. Many steps can be done naturally in your mind.

Or: 12 : 81+108 12 : 189 _ all divided by 3 4 : 63

other way to solve the equation y³ + y² - 2/27 = 0 1. Find a solution by inspection. I found one after trying three fractions 2. the solution ist y = - 1/3 3. factorize y³ + y² - 2/27 = ( y + 1/3 )( y² + ay - 2/9 ) 4. the factorizing can be found by long division --> a =+2/3 5. solving ( y + 1/3 )( y² + 2y/3 - 2/9 ) =0 6. solution 1: y = - 1/3 isn't a solution, while y must be positive 7 solution 2: y =- (-1-root(3))/3 isn't a solution, while y must be positive 8. solution 3: y -= (-1+root(3))/3 is a solution, while y is positive 9. for y ^6 calculate {(-1+root(3))/3]² * {(-1+root(3))/3]² * {(-1+root(3))/3]²

Math is life