Curious why you decided to solve for y instead of x when given the option? Feel like this was a lot of unnecessary work

3^1 2^2 3^1 1^2 32 (x ➖ 3x+2) .

x⁵+x⁴+x³+x²+x+1=0 (1) x⁶+x⁵+x⁴+x³+x²+x =0 (2) Substract (1) from (2): x⁶-1=0 (x³+1)(x³-1)=0 (x+1)(x²-x+1)(x-1)(x²+x+1)=] x+1=0 --> x=-1 x²-x+1=0 --> x=½[1±isqrt(3)] x-1=0 --> x=1 x²+x+1=0 --> x=½[-1±isqrt(5)]

x is exactly 8 .

Excellent. But where did you get alpha from

Let a/b = z. Then z+1 = 6*sqrt (z) --> (z+1) ^2 = 6^2 * z -> z^2 + 2z + 1 = 36z -> z^2 - 34z + 1 = 0 -> z^2 - 34z + 289 = 288 -> (z-17) ^2 = 2*14^2 -> z=a/b = 17 +/- 12*sqrt (2) [I.e., there are 2 solutions]

I disagree with your solution, sir. I have X = ±√(7+√13) and X = ±√(7-√13). They're supposed to be four roots

Squring Both Sides. We get a^2+b^2+2ab=36ab so a^2+b^2=34ab. We DIVIDE Both Sides with b^2>0, hence (a^2/b^2)+(b^2/b^2)=34ab/b^2 so (a/b)^2+1=34*(a/b). Setting a/b=x we get x^2+1=34x so x^2-34x+1=0 so x=(34+ -sqrt(34^2-4))/2. Using trick of Difference of squares 34^2-2^2=(34+2)*(34-2)=36*32 so x= (34+ -6*sqrt32)/2= 17+ - 3sqrt32= 17+ - 3*sqrt16*sqrt2=17+ -12*sqrt2.

Easier to divide original equation by b giving: x² + 1 = 6 sqrt(x). Square both sides gives: x² - 34x +1 = 0 Simple.

I've found the shortcut for such assignment: (a^3 + b ^3) /(a^3 + c^3); where a = b + c (a^3 + b ^3) /(a^3 + c^3) = (a +b) / (a +c) or (a^3 + c ^3) /(a^3 + b^3) = (a +c) / (a +b) for a = 97; b = 54; c = 43 (97^3 + 54 ^3) /(97^3 + 43^3) = (97 + 54) / (97+ 43) = 151/140

43^54^3^1+22^32^3^1/43^54^3^1+43^3^1 1^1^1 +2^11^2^161^1/1^1^1^1 1^1 1^1^1 1^4^4 / 2^2^2^2/ 1^11^2 1^2 (x ➖ 2x+1)

Thank uu

♥

φ = (1+√5)/2 φⁿ = F(n)φ + F(n-1), F(n) Fibonacci [proof is trivial by induction] φ¹²=F(12)φ + F(11) = 144φ + 89 = 161 + 72√5

(1^1+5^1/2^1)3^4(1^1/1^1)3^2^2 (1^1)3^2^1^2 3^1^1^2 3^2 (x ➖ 3x+2)

(1 + R5)^3 = 1^3 + 3 R5 + 3 (R5)^2 + R5^3 = 1 + 3 R5 + 3 * 5 + 5 R5 = 16 + 8 R5, hence ((1+R5)/2)^3 = 2 + R5. (2+R5)^2 = 4 + 2 * 2 R5 + 5 = 9 + 4R5. (9 + 4R5)^2 = 81 + 9 * 4 * 2 R5 + 16 * 5 = 161 + 72 R5. Not such a nice solution, but only need binomal formulas for squaring and raising to the power 3, which I consider standard for 16 year old high school pupils

63+26=89

161+72√5

because 1/(x(x+4))=(1/4)(1/x-1/(x+4)), then 1/5+1/45+1/117+1/221+1/357+1/525 =(1/4)(1/1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21+1/21-1/25) =(1/4)(1/1-1/25) =6/25

Good simplification sir

It is queer that 221 = 13 x 7. But I know that 13 x 7 = 91 So great an error is swooning

Clear the fraction and let a=x^2-6 and b=x. You'll get a much simpler problem. The solution will be a nested root but will de-nest nicely.

*=read as square root 1/5=1/(1×5)=1_(1/5)=1-(1/5)=4/5 Similarly 1/45=1/(5×9)=1/5 - 1 /9=4/45 1/525=1/21 -1/25 The given series will be 1/4(1- 1/25) 1/4(24/25)=6/25 *(6/25)=*6/5

n=20

Use 1/(9x5)=(1/5-1/9)/4 and 1/(9x13)=(1/9-1/13)/4, .... You will end up = (1-1/25)/4 = 6/25.

Math Olympiad? for 1st grader?

Another approach : x/21 + x/77 + x/165 + x/285 = 200 x/(3*7) + x(7*11) + x(15*11) +x (15*19) We can use that trick: Observe that the difference between both factors in each denominators is 4 (it's our D) so, we can use the multiplication of 1/D by the difference between the smallest factor and the biggest one divided by product of ab. a = 3; b = 19 ------> (b - a) / (ab) and D = 4 ---------------> 1/D 1/D [(b-a)/ab x/4(19 - 3)/ 3*19 = 200 x/4(16/57) = 200 4x / 57 = 200 x = 50*57 x = 2850 This very useful concept is based on identity: 1/a - 1/b = b-a/ ab and 1/ab = [(1/a) - (1/b)] (1/ (b-a)

this also proves 36=16 if you expand (4-5)^2=(6-5)^2 WOAH!!!

No substitution (x²-x-6)²+(x²+x-6)²=6x² (x²+x-6-x²+x+6)²+2(x²+x-6)(x²-x-6)=6x² 4x²+2(x²-4)(x²-9)-6x²=0 ÷2 x⁴-14x²+36=0 ⇒(x²-7)²=13 x=±sqr(7±sqr13) By the way your voice is similar to affrican people You from affrica? My greeting to you

17/9=8.1 17/4=4.1 {8.1+4.1}= 12.2 3^4 2^1 3^2^2 .2^1 3^1^2.1^1 3^2 ( y ➖ 3x+2)

Ustad from India. ❤

Yes. He did. The other roots are x≈±1.84 thus x≈±3.26 😅

(6^(1/2))/5.

root a^2 = +-a; ya went wrong there...

You can stand on your head your head and whistle through your backside 1+1can never be 3.neither can you divide by 0. Don't teach wrong things. Why don't you use sandpaper instead of toilet paper 😂😂😂😂

Bhoat acha. You much be from India.

Thanks for writing down the rules of exponents. We must be really dumb

Magic. So cool

Thank you for explaining. I have a question. If we use " f(x)≧0 for √f(x) ", x=2 should be rejected because all values "inside of the √ " are negative. If x=2 is one of the solutions, I guess the problem admits complex numbers. (: The reason of rejection is not "Not Real.") ・・・ If x=2, (the left side) = 3√2 i + 2√2 i and (the right side) = 5√2 i . Thus, it can satisfy the equation (, but it is not real numbers.) And, even if the complexed numbers are admitted, if x=-5/3, (the left side) = (4√17)/3 i + (√17)/3 i and (the right side) = (√17) i . Thus, it does NOT satisfy the equation. Therefore, I guess the reason why x=-5/3 is rejected in this video is a little different. <<<<< Anyway, if the rejecting reason is "Not Real", x=2 should be rejected (as well as x=-5/3). On the other hand, if the rejecting reason is "Not Satisfy the given Equation" ( but complex number is OK ), only x=-5/3 is rejected. >>>>> (My question is "which condition is the original problem requiring?" )

Interesting question, thank you very much for sharing 🙏👌

Philippine mathematical Olympiad: [(x² - x - 6)/6x]² + [(x² + x - 6)/6x]² = 1/6 x ≠ 0, [(x² - 6) - x]² + [(x² - 6) + x]² = (6x)²(1/6) = 6x², 2[(x² - 6)² + x²] = 6x² (x² - 6)² - 2x² = 0, (x² - 6)² - [(√2)x]² = 0, [x² - 6 - (√2)x][x² - 6 + (√2)x] = 0 x² - (√2)x - 6 = 0 or x² + (√2)x - 6 = 0 x = (√2 ± √26)/2 or x = (- √2 ± √26)/2 Answer check: [(x² - x - 6)/6x]² + [(x² + x - 6)/6x]² = 1/6 x = (√2 ± √26)/2: x² - (√2)x - 6 = 0, x² - 6 = (√2)x [(√2 - 1)² + (√2 + 1)²]/36 = (3 + 3)/36 = 1/6; Confirmed x = (- √2 ± √26)/2: x² + (√2)x - 6 = 0, x² - 6 = - (√2)x [(- √2 - 1)² + (- √2 + 1)²]/36 = [(√2 + 1)² + (√2 - 1)²]/36 = 1/6; Confirmed Final answer: x = (√2 + √26)/2; (√2 - √26)/2; (- √2 - √26)/2 or (- √2 - √26)/2

German math Olympiad: 5^(x² - 6x + 7) = 0.64; x = ? Let: 5ª = 0.64, a = log₅(0.64) = - 0.277, 5^(x² - 6x + 7) = 5^(- 0.277) x² - 6x + 7 = - 0.277, (x - 3)² = - 7.277 + 9 = 1.723 x = 3 ± √1.723 = 3 ± 1.313; x = 3 + 1.313 = 4.313 or x = 3 - 1.313 = 1.687 Answer check: x² - 6x + 7 = log₅(0.64) = - 0.277 x = 4.313 x² - 6x + 7 = x(x - 6) + 7 = 4.313(4.313 - 6) + 7 = - 0.276; Confirmed x = 1.687 x² - 6x + 7 = 1.687(1.687 - 6) + 7 = - 0.276; Confirmed Math calculation was achieved on a smartphone with a standard calculator app Final answer: x = 4.313 or x = 1.687

Indian Math Olympiad: (28 + 10√3)ˣ + (5 + √3)ˣ = 12; x = ? 28 + 10√3 = 25 + 2(5)(√3) + 3 = 5² + 2(5)(√3) + (√3)² = (5 + √3)² Let: a = 5 + √3 = 6.732, (28 + 10√3)ˣ + (5 + √3)ˣ = a²ˣ + aˣ = 12 (aˣ)² + aˣ - 12 = (aˣ + 4)(aˣ - 3) = 0, aˣ = 5 + √3 > 0, aˣ + 4 > 0 aˣ - 3 = 0, aˣ = 3, x = logₐ3 = (log3)/(log6.732) = 0.576 Math calculation was achieved on a smartphone with a standard calculator app Answer check: (28 + 10√3)ˣ + (5 + √3)ˣ = a²ˣ + aˣ = aˣ(aˣ + 1); a = 5 + √3 x = logₐ3 = 0.576, aˣ = 3: aˣ(aˣ + 1) = 3(3 + 1) = 12; Confirmed Final answer: x = logₐ3 = 0.576

Ψ = Χ^2 - 6, [ (Ψ +Χ) / 6Χ ] ^2+ [(Ψ-Χ) / 6Χ ] ^2 = 1 / 6 [(Ψ+Χ) / 6Χ] - [(Ψ-Χ) / 6Χ] = (Ψ+Χ-Ψ+Χ) / 6Χ = 2Χ / 6Χ = 1 / 3 {[(Ψ+Χ) / 6Χ] - [(Ψ-Χ) / 6Χ]} ^2 = (1/3) ^2 [(Ψ+Χ) / 6Χ] ^2 + [(Ψ-Χ) / 6Χ] ^2 - 2 [(Ψ+Χ)(Ψ-Χ)] / 36Χ^2 = 1 / 9 1/6 - (Ψ^2 - Χ^2) / 18Χ^2 = 1 / 9 1 / 6 - 1/9 = (Ψ^2-Χ^2) / 18Χ^2 1 / 18 = (Ψ^2-Χ^2) / 18Χ^2, Ψ^2-Χ^2 = Χ^2 Ψ^2 = 2Χ^2 Ψ = Χ. 2^1/2 , Ψ = -Χ. 2^1^2 Χ.2^1/2 = Χ^2 - 6 Χ = (2^1/2 + 26^2)/2 Χ = (-2^1/2 - 26^1/2)/2

Where is solution when a,b are not integer?

a=24 and b =96

Делим на х и умножить на 36 (х-6/х-1)^2+(х-6/х+1)^2=6 Легко видно что а=х-6/х Тогда (а+1)^2+(а-1)^2=6 а^2+1=3 а=sqrt 2 или а= - sqrt 2 Дальнейшее просто Два простых квадратных уравнений х-6/х-sqrt2=0 или х-6/х+sqrt 2=0 Решение которых не составит труда Решение от силы 3 - 4 минуты, но не 10 как на видео. Если это часть теста, то вы потеряли 6 минут просто так

2* (x^2 - 6) ^2 + 2 * (x )^2 = 6 x^2 ( x^2 - 6)^2 = 2 x^2 ( x^2 - √2 x - 6) ( x^2 + √2 x - 6) = 0 (( x - 1/√2) ^2 - 13/2) * (( x + 1/√2) ^2 - 13/2) = 0 x = (1 + √(13)) /√2, (1-√(13)) /√2, -(1 + √(13)) /√2, (-1+√(13)) /√2,

Horrible video. You would have obtained a polynomial of degree 4 and you only found 2 solutions. You cannot disregard the plus or minus sign in this case.

A fairly easy question. I solved it just in my head by using the natural logarithm after the factorization.It will give us the same result and in a more direct way n³=-1. But, first the question does not specify whether n is a complex or a real number. Because in the set of real numbers there exists a unique solution which is -1. But in the set of complexe numbers there exist 3, which are -1,-j,-j² in this case, as it is -1 and not 1 (the negative of the cubic root of unity of a complexe number). In any case good exercise