Beautiful

Another? g1= x^4+y^4-68; g2=x*y*(2*(x^2+y^2)+3*x); (x+y)^4 = (x^4+y^4)+2*x*y*(2*(x^2+y^2)+3*x*y) =68+2*128 = 324 =4*3^4. => (x+y)= +/- 3*sqrt(2). Now define g3=x+y-3*sqrt(2) (=0). g4=x+y+3*sqrt(2) (=0); now eliminate 'x' by dividing g1/g3 to obtain P1(y). Then divide g1/g4 to obtain P2(y). where P1(y)=0 &P2(y)=0. Here P1(y)=2*y^4 - 12*2^(1/2)*y^3 + 108*y^2 - 216*2^(1/2)*y + 256 P2(y) = 2*y^4 + 12*2^(1/2)*y^3 + 108*y^2 + 216*2^(1/2)*y + 256. P1(y)=0 => y={ [2^(1/2), 2*2^(1/2), (3*2^(1/2))/2 +/- (110^(1/2)*1i)/2 }, P2(y)=>y={-2^(1/2), -2*2^(1/2), - (3*2^(1/2))/2 + (110^(1/2)*1i)/2}

2y^2/y^2 =2y^1 (y ➖ 2y+1).{2x^2/x^4+22x^2/xy^8}=24x^4/xxy^12 =2xxy^3 (xxy ➖ 3xxy+2).

{6^6x^2 +6}= 6^12x^2 3^2^3^4x^2 1^3^2^2x^1 ^3^2^1^2x 3^1^1^2x 3^2x (x ➖ 3x+2).3^2x^2+ {6x^2+1}/2= 7x^2/2 {3^2x^2+7x^2/2} = {10x^4/2}= 5x^2 5^1x^2 1^1x^2 1x^2 (x ➖ 2x+1).

it's a cubic equation 3ˣ = u u³ + 3 = u(3 + √3) ..... I don't know how to continue .....

Εχω τελικα (χ-y)(χ+y+χy)=11. Αλλα 11=1×11ή 11×1 ή-1×(-11) ή -11×(-1). Τελικα (χ,y)=(-4,-5) (3,2) (-1,10) (-12,-1)

Αν y=3^χ >0 τοτε χ=1/2 στο Q ενω στο R επι πλεον και χ=[log(-ριζα3+ριζα(3+4ριζα3))-log2]/log3

Let 3^x=t. Then, t^3-(3+√3)t+3=0. So t=√3 is a solution > x=1/2. Again, if x is real, t>0. Now, [t^3-(3+√3)t+3]/(t-√3)= t^2 +√3t-√3 and t^2 +√3t-√3=0 > t = 1/2[√(3+4√3) -√3] and hence x = 1/(ln3) ln {1/2[√(3+4√3) -√3] }. So, x = 1/2, 1/(ln3) ln {1/2[√(3+4√3) -√3] }.

alternative? x &y simplify to x= 5+2*sqrt(6), y= 5-2*sqrt(6), also x^5-y^5 = (x-y)*( (x^4+y^4) + x*y*(x^2+y^2) + (x*y)^2) ) x - y = 4* sqrt(6), , x + y = 10, x*y =1=> x^5-y^4 = 4* sqrt(6)*( (x^4+y^4) + (x^2+y^2) + 1 ) (x+y)^2 = 100 =(x^2+y^2 +2) => x^2+y^2 = 98, (x+y)^4 = ( (x^4+y^4) +98+1)=> x^4+y^4 = 9602=> x^5-y^5 = 38804*sqrt(6)

(x,y)=(3,2), (-4,-5), (-1,10), (-12,-1)

(x,y)=(3,2) also other solutions

X=1/2, log(root (3+4√3)-√3)/log3 ,log(root(3-4√3)-√3)/log3

X=0.5

Comparing well on both sides give x= 1/2 the soln.

x^2(x^4 + 1) + 6(5x + 24) = 2x^3(x + 15) x^6 + x^2 + 30x + 144 = 2x^3(x + 15) x^6 + (x + 15)^2 - 81 = 2x^3(x + 15) {x^3 - (x + 15)}^2 - 9^2 = 0 (x^3 - x - 24)(x^3 - x - 6) = 0

If x-9=t, x =t+9. Thus, t+9 -93/(t+9) = -1 > t-3/t = -19 > t/(sqrt3) - (sqrt3)/t = -19/sqrt(3) > t^2/3+3/t^2 = 367/3 > t^4/9 + 9/t^4 = (367/3)^2 - 2 > The required expression is 134671/9.

It was a wonderful introduction and clearly explained...thanks Sir for sharing 🙏.....E = 134671/9

2^(2x) = u 4³u³ = (1 - 8u)²/4² 4⁵u³ = 8²u² - 16u + 1 2¹⁰u³ - 2⁶u² + 2⁴u - 1 = 0 2⁶u²(2⁴u - 1) + (2⁴u - 1) = 0 (2⁴u - 1)(2⁶u² + 1) = 0 2⁴u - 1 = 0 => u = 1/2⁴ 2^(2x) = 1/2⁴ 2x = -4 => *x = -2* 2⁶u² + 1 = 0 => no real solutions

My method shows I have no talent

x^2-2*x-93=0>x= -1/2 +/-sqrt(373)/2. define Z(1) =(x-9)/sqrt(3) + sqrt(3)/(x-9). (where Z(n) = (Y)^n +1/(Y)^n ). Now use standard recursion for Z(n), i.e. Z(n+1)= Z(n)Z(1) - Z(n-1) & Z(2n) = Z(n)^2-2. Here only need the doubling formula. i.e. Z(2n) = Z(n)^2-2. Here Z(1) = sqrt(1119)/3 => Z(2)= Z(1)^2-2 = 367/3 => Z(4) = Z(2)^2 -2 = 134671/9

I solved this as follows: Let t=2(x+1)-6/(x+1), then the given equation is (t-1)^2+(t+1)^2=4⇔t=±1. 2(x+1)-6/(x+1)=±1 ⇔2x^2+3x-5=0 or 2x^2+5x-3=0 It is intriguing that x's such that either of the numerators is zero are solutions.

(x)^2 ➖ 93/(x)^2= {x^2 ➖ 93}/x^2 =91/x^2 =92^ 62^31 2^31^31^1 2^31^1 1^1 2^1^1 2^1 (x ➖ 2x+1). { 6561x^4/9+ 9/6561x^4 } {6570x^4/6570x^4} =1x^1 (x ➖ 1x+1).

{8x^2+9x^2➖} (5)^2)=| 17x^4 ➖ 25}》= 8x^4/2x^2 =4x 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1) {8x^2+10x^2} (3)^2 ={18x^4 ➖ 9} =9x^4 /2x^2 =4 1x^2 2^2.1^1x^2^1.1^1x^2^1 x^2^1 (x ➖ 2x+1).

Marathon problem! Nice choice and nicely solved - even though the final result isn't so pretty! Thank you.

Υπέροχπο θέμα μπράβο

x - 9 = u => x = u + 9 E = u⁴/9 + 9/u⁴ = (u/√3)⁴ + (√3/u)⁴ u + 9 - 93/(u + 9) = -1 (u + 9)² + (u + 9) - 93 = 0 u² + 18u + 81 + u + 9 - 93 = 0 u² + 19u - 3 = 0 (u² + 19u - 3)/(u√3) = 0 u/√3 + 19/√3 - √3/u = 0 u/√3 - √3/u = -19/√3 (u/√3)² + (√3/u)² = 19²/3 + 2 (u/√3)⁴ + (√3/u)⁴ = (19²/3 + 2)² - 2 E = (19²/3 + 2)² - 2 E = 19⁴/9 + (4)19²/3 + 2 *E = (19⁴ + (12)19² + 18)/9* *E = 134671/9*

E=134671/9

2/3

x=-3, x=-5/2, x=1/2 and x=1

X=1, -3, 1/2; -5/2.

By letting (2x² +3x -5)/(x +1) =y the given becomes y² + (y +2)² = 4, and then y² + y² +4y +4 = 4; 2y² +4y +4 =4; 2y² +4y =0; y² +2y =0; y(y+2) =0, that is, y =0 or y =-2 Thus, recalling 'y' (2x² +3x -5)/(x +1) =0 (e1) or (2x² +3x -5)/(x +1) =-2 (e2) Solving (e1) 2x² +3x -5 =0; (x -1)(2x +5) =0; x = 1, -5/2 Solving (e2) 2x² +3x -5 =-2(x +1); 2x² +5x -3 = 0; (x +3)(2x -1) =0; x = -3, 1/2 Over all, x = -3, -5/2, 1/2, 1

Θετωy=2(x)^2+4x-4. και εχω y/(x+1)=+ -1 αρα x=-5/2, 1, -3, 1/2

X=1; -5/2; 1/2; -3 Ironically; both upper terms of l. H. S are the solns.

Note that [2x^2+5x-3]/(x+1) = [2x^2+3x-5]/(x+1 +2. So, let t= [2x^2+3x-5]/(x+1) +1 = [2x^2+4x-4]/(x+1). Then the given equation becomes (t+1)^2 + (t-1)^2 = 4 > t^2=1 > t = +/-1. If t=1, 2x*2+3x-5=0 > x= -5/2, 1. If t=-1, 2x^2+5x-3=0 > x=-3, 1/2. Thus, x=-3, -5/2, 1/2, 1.

Amazing!

A Nice Exponential Equation: √[(2^(2x + 5)/8] = ³√{[1 - 2^(2x + 3)]/4}, x ϵR √[(2^(2x + 5)/8] = √[(2²ˣ)(2⁵)/8] = √[(2²)(2ˣ)²] = 2(2ˣ) [1 - 2^(2x + 3)]/4 = [1 - 8(2ˣ)²]/4 = [2(2ˣ)]³ = 8(2ˣ)³, Let: y = 2ˣ (1 - 8y²)/4 = 8y³, 32y³ + 8y² - 1 = 0, y³ + (1/4)y² - 1 /32 = 0 (y³ - 1/64) + [(1/4)y² - 1/64] = [y³ - (1/4)³] + (1/4)[y² - (1/4)²] = 0 (y - 1/4)(y² + y/4 + 1/16) + (1/4)(y + 1/4)(y - 1/4) = 0 (y - 1/4)(y² + y/4 + 1/16 + y/4 + 1/16) = (y - 1/4)(y² + y/2 + 1/8) = 0 y = 2ˣ > 0, x ϵR: y² + y/2 + 1/8 > 0; y - 1/4 = 0, y = 2ˣ = 1/4 = 2⁻², x = - 2 Answer check: x = - 2: 2ˣ = 1/4 √[(2^(2x + 5)/8] = 2(2ˣ), ³√{[1 - 2^(2x + 3)]/4} = ³√{[1 - 8(2ˣ)²]/4} 2(2ˣ) = 2(1/4) = 1/2, ³√[(1 - 1/2)/4] = ³√(1/8) = 1/2; Confirmed Final answer: x = - 2

Самая лучшая замена а=2^(2х+3) Sqrt (a/2)=sqrt (3)((1-a)/4) Возводим в 6 степень а^3/8=(1-а)^2/16 2*а^3-а^2+2*а-1=0 2а(а^2+1)-(а^2+1)=0 2а-1=0, а=2^(-1)=2^(2х+3) 2х+3=-1 х=-2 3 минуты, а не 12 как на видео

Chase out the perfect squares!!! That's pretty fundamental for a final answer, giving 38804 * sqrt(6).

Easier? g1=x+4-(y-2)^2 (= 0), g2=y+4-(x-2)^2 (=0), g3=x^2+y^2 (=?). 1st step g2/g1 -> p1= - y^4 + 8*y^3 - 12*y^2 - 15*y (=0) 2nd step g3/g1 -> p2= y^4 - 8*y^3 + 17*y^2 (=?). 3rd step p2/p1 -> p3 = 5*y^2 - 15*y (= x^2+y^2). Now solve p1 to obtain values, 'y' p1= -y*(y-5)*(y^2-3*y-3) => y={ 0, 5, 3/2 - 21^(1/2)/2, 21^(1/2)/2 + 3/2} . Subs y into p3 =>x^2+y^2 = {0, 50, 15, 15}

(20/96+{(1)^2/*8)^2 ➖ (1)^2/(12)^4 }= {1/64 ➖ 1/244} ={0+0 ➖}/180.=1/180 {20/96+1/180}=21/276 =10.66 2^5.6^11 2^1.6^11^1 2^1.61^1 2^1.3^2^1^1 1^1.3^2 3^2 (x ➖ 3x+2). (1)^2/(24)^4 = 1/1152{ 1+1 ➖/8+8 ➖ +1+1 ➖/12+12 ➖} =4/40 {1/1152 ➖ 4/40} =3/1112 =37.2 37^1.2^1 1^1 .2^1 2^1(x ➖ 2x+1)

let g1 =a+b-1 (=0), g2=a^6+b^6-65 (=0). Since g1 is 1st order in 'a' and 'b' we can use synthetic division to eliminate one of the components for example using 'a' as reference then remainder g2/g1 = p(b)=2*b^6 - 6*b^5 + 15*b^4 - 20*b^3 + 15*b^2 - 6*b - 64; p(b) = (b - 2)*( b + 1)*( 2*b^4 - 4*b^3 + 15*b^2 - 13*b + 32) so (b=2, a=-1), ( b=-1, a=-2 )

8x^2^10/8=1.2x^1^.2 1^12^1x1^12^1 1^1'x^2^1 x^2^1 (x ➖ 2x+1) (1)^2 ➖ (2)^2{2x+2x ➖ 3+3 ➖ }{ 1 ➖ 4}{4x^2+6} ={3+10x^2}/4 1=13x^2/4=3.1x^2 3^1.1^1x^2^1 3x^2 (x ➖ 3x+2)