Let x = 1+t , y = 1- t 2(1+6 t^2+t^4) = 82 t^4 + 6 t^2 - 40 = 0 t^2 = 4 , -10 t = 2 , - 2 Case 1 , x = 3 , y = - 1 Case 2 , x= -1 , y = 3

И для какого класса в Америке это уравнение ?? 🤔🤔🤔🤔 В России это уровень 7 класса

x² + 3x = 10000099998 x² + 3x + (3/2)² = 10000099998 + (3/2)² [x + (3/2)]² = 10000099998 + (9/4) [x + (3/2)]² = 10000000000 + 99998 + (9/4) [x + (3/2)]² = 100000² + 99998 + (9/4) [x + (3/2)]² = 100000² + 99998 + 2 - 2 + (9/4) [x + (3/2)]² = 100000² + 100000 - 2 + (9/4) [x + (3/2)]² = 100000² + 100000 + (1/4) [x + (3/2)]² = 100000² + 100000 + (1/2)² [x + (3/2)]² = 100000² + 2.[100000 * (1/2)] + (1/2)² [x + (3/2)]² = [100000 + (1/2)]² x + (3/2) = ± [100000 + (1/2)] First case: x + (3/2) = [100000 + (1/2)] x + (3/2) = 100000 + (1/2) x = 100000 + (1/2) - (3/2) x = 100000 - (2/2) x = 100000 - 1 x = 99999 Second case: x + (3/2) = - [100000 + (1/2)] x + (3/2) = - 100000 - (1/2) x = - 100000 - (1/2) - (3/2) x = - 100000 - (4/2) x = - 100000 - 2 x = - 100002

With respect, there is no need to find a and b since, (ab)^2=27^2=45^2-x^2, hence, x^2=45^2-27^2=(45-27)(45+27) =(18)(72)=9(2)9(8)=9^2x4^2, hence, x=9(4)=36 or -36

10000099998 = 3 * 3333366666 10000099998 = 3 * 33333 * 100002 10000099998 = 99999 * (99999+3) 10000099998 = 99999^2 + 3 * 99999 x^2 + 3 x = 99999^2 + 3 * 99999 x = 99999

10000099998 = x^2 + 3 x 10000099998 + 2 = x^2 + 3 x + 2 10000100000 = (x+1)(x+2) 100000*100001= (x+1)(x+2) (99999+1)(99999+2)= (x+1)(x+2) x = 99999

40=49-9 =[sqrt(7)]⁴-[sqrt(3)]⁴ Therefore sqrt(x)=4 --> x=16

Sqrt[((45+x)^3)/(45-x)]+Sqrt[((45-x)^3)/(45+x)]=246 x=±36 x=±15Sqrt[91]i

(135+x^3/(45 ➖ x)^2+(17 275 ➖ x^3)/(135+x^3)=135x^3/(2025 ➖ x^2)+{x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖ }/135x^3=135+x^3/{x^0+x^0 ➖ x^0+x^0 ➖}+x^3/135x^3 ={135x^3/x^2+x^3/135x^3}=135x^6/135x^5=1x^1.1 1x^1 (x ➖ 1x+1).

Yo habría usado el mínimo comun multiplo y habría resultado la ecuación como una fraccion normal.....si después me queda una ecuación de 4° grado uso Ruffini y listo....no es lo mas rápido, pero si lo mas efectivo.

For which class? If for the elder, then prove that the function 7^x-3^x increases monotonously - if you want algebraically, you want through a derivative. And the answer 49-9=40 is obvious.

Multiply eqn with(√45^ 2- x^2) gives 45^2+ x^2= 126(√45^2-x^2) let a= √(45^ 2- x^2)gives a^2+ 123a- 4050= 0 4050= 150× 27 hence a= 27; -150 a> 0 so only 27 valid. Hence x^2= 45^2 - 27^2= 72×18= (36)^2 or X = + -36 solns.

Sqrt[7]^Sqrt[16]-Sqrt[3]^Sqrt[16]=40 X=16 final answer

4:53 How do you derive a and b are integers?

Argentina, Math Olympiad: (√7)^√x - (√3)^√x = 40, x ϵ ℤ; x =? (√7)^√x - (√3)^√x = 40 = 49 - 9 = 7^2 - 3^2 = (√7)^4 - (√3)^4 = (√7)^√16 - (√3)^√16 x = 16

{49 ➖ 9}=40 2^20 2^10 2^2^5 1^2^1 2^1 (x ➖ 2x+1).

{9x+9x ➖ }+{4x+4x ➖ }+{4x+4x ➖ }+{2x+2x ➖ }+{3x+3x ➖ }={18x^2+8x^2+8x^2+4x^2+6x^2}=40x^10 ^18^8^8^4^6x^10 2^9^2^3^2^3^4^2^3x^5^5 1^3^2^1^1^1^1^1^1^1^1x^2^3^2^3 1^1x^2^1^1^3 x^2^3(x ➖ 3x+2).

49- 9= 40 gives √ x = 4 hence x= 16 +z

Nice. I used trial and error lol. And found 5.

√{9 + 4√(4 + 2√3)} = √{9 + 4√(3 + 2√3 + 1)} = √{9 + 4√(√3² + 2√3 + √1²)} = √{9 + 4√(1 + √3)²} = √{9 + 4(1 + √3)} = √(9 + 4 + 4√3) = √(13 + 2√4*3) = √(1 + 2√12 + 12) = √(√1² + 2√12 + √12²) = √(1 + √12)² = 1 + √12 = 1 + 2√3

Nice radical simplification problem in double deck. (1 + 2√3) is simplified soln.

Then u will get x^2+3x = t^2+3t So x= t = 9999 Also x=-t-3=-10002

Put 9.1111 =t

Very simple to solve. Complete the square at the left hand side by adding (³⁄₂)² = ⁹⁄₄ = 2 + ¹⁄₄ to both sides, then we have x² + 3x + ⁹⁄₄ = 10¹⁰ + 10⁵ + ¹⁄₄ (x + ³⁄₂)² = (10⁵ + ¹⁄₂)² x + ³⁄₂ = 10⁵ + ¹⁄₂ ⋁ x + ³⁄₂ = −10⁵ − ¹⁄₂ x = 10⁵ − 1 ⋁ x = −10⁵ − 2

³√(x + 3) = √(x - 1) [³√(x + 3)]³ = [√(x - 1)]³ (x + 3) = (x - 1) √(x - 1) (x + 3)/(x - 1) = √(x - 1) [(x + 3)/(x - 1)]² = [√(x - 1)]² (x + 3)²/(x - 1)² = (x - 1) (x + 3)² = (x - 1)³ Put, y = x - 1 (y + 4)² = y³ y³ - y² - 8y - 16 = 0 Put, F(y) = y³ - y² - 8y - 16 F(4) = 64 - 16 - 32 - 16 = 0 ∴ [ y³ - y² - 8y - 16 = 0 ] is must have a factor "4" y³ - y² - 8y - 16 = 0 y³ - 16y - y² + 8y - 16 = 0 y(y² - 16) - (y² - 8y + 16) = 0 y(y + 4)(y - 4) - (y - 4)² = 0 (y - 4)[y(y + 4) - (y - 4)] = 0 (y - 4)(y² + 4y - y + 4) = 0 (y - 4)(y² + 3y + 4) = 0 y - 4 = 0 , y² + 3y + 4 = 0 ∴ y = 4 <--- y = x - 1 x - 1 = 4 ∴ x = 5 y² + 3y + 4 = 0 D = 9 - 4*1*4 = 9 - 16 = -7 < 0 y = (- 3 ± √D) / 2*1 = (- 3 ± √-7) / 2 ∴ y = (- 3 ± √7i) / 2 <--- y = x - 1 x - 1 = (- 3 ± √7i) / 2 x = (- 3 ± √7i) / 2 + 1 ∴ x = (- 1 ± √7i) / 2

x² + 3x = 10,000,099,998 x² + 3x - 10,000,099,998 = 0 D= 9 + 4*10,000,099,998 = 40,000,400,001 = 200,001² x = (- 3 ± √200,001²) / 2*1 = (- 3 ± 200,001) / 2 x = 199,998 / 2 or - 200,004 / 2 ∴ x = 99,999 or x = - 100,002 10,000,099,998 = 10,000,000,000 + 99,998 = 100,000² + 99,999 - 1 = (99,999 + 1)² + (99,999 - 1) = (9*11,111 + 1)² + (9*11,111 - 1) x² + 3x = (9*11,111 + 1)² + (9*11,111 - 1) Put, a = 11,111 x² + 3x = (9a + 1)² + (9a - 1) x² + 3x = 81a² + 18a + 1 + 9a - 1 x² + 3x = 81a² + 27a x² - 81a² + 3x - 27a = 0 (x + 9a)(x - 9a) + 3(x - 9a) = 0 (x - 9a)(x + 9a + 3) = 0 x - 9a = 0 , x + 9a + 3 = 0 <--- a = 11,111 x = 9a = 9 * 11,111 = 99,999 x = - 3 - 9a = - 3 - 99,999 = -100,002

{x^3+x^3 ➖}+{x+x ➖}+{3+3 ➖ }{x^6+x^2+6} 6x^8 6x^2^3 2^3x^1^1^1 2^3x (x ➖ 3x+2).(x)^2 ➖ (1)^2 {x^2➖ 1}={x^0+x^0 ➖ x^0+x^0 ➖ }={x^1+x^1}=x^2 (x ➖ 2x+2).

{x^2+x^2 ➖ }+{3x+3x ➖ }={x^4+6x^2}6x^6 2^3x^2^3 2^1x1^3 2x^3 (x ➖ 3x+2).1^3^2^3^2^3^2^3^2^2^3 1^1^1^1^1^1^1^1^3^1^1^2 32(x ➖ 3x+2).

14^2=196

On raising the power of 6 B/S : x2 + 6x + 9 = x^3 - 3x^2+ 3x - 1 x^3 - 4x^2 -3 x -10= 0 (x - 5) (x^2 + x + 2) = 0 x = 5 plus two roots of complex category

Shorts....r. h. S = 9( 1111122222)= 9× 11111(100002)= 99999( 100002) now obvios solns x= 99999; -100002

Excelente explanação.

R U kidding me? (2¹⁰¹ + 8) / (16¹³ + 4¹⁴ + 8) = 2(2¹⁰⁰ + 2²) / (2²*2⁵⁰ + 2³*2²⁵ + 2*2²) = 2(2¹⁰⁰ + 2²) / 2(2*2⁵⁰ + 2²*2²⁵ + 2²) = (2¹⁰⁰ + 2²) / (2*2⁵⁰ + 2²*2²⁵ + 2²) = {(2⁵⁰)² + 2²} / 2(2⁵⁰ + 2*2²⁵ + 2) Put, t = 2²⁵ {(t²)² + 2²} / 2(t² + 2t + 2) = {(t² + 2)² - 4t²} / 2(t² + 2t + 2) = (t² + 2t + 2)(t² - 2t + 2)/ 2(t² + 2t + 2) = (t² - 2t + 2)/ 2 <--- t = 2²⁵ = {(2²⁵)² - 2(2²⁵) + 2}/ 2 = {(2*2²⁴)² - 2(2*2²⁴) + 2}/ 2 = {2²(2²⁴)² - 2²*2²⁴ + 2}/ 2 = 2(2²⁴)² - 2*2²⁴ + 1 = 2^49 - 2^25 + 1 2(2²⁴)² - 2*2²⁴ + 1 = 2(2⁴*2²⁰)² - 2*2⁴*2²⁰ + 1 = 2^9*(2²⁰)² - 2⁵*2²⁰ + 1 = 512*(2²⁰)² - 32*2²⁰ + 1 = 512*(2¹⁰)⁴ - 32*(2¹⁰)² + 1 <--- 2¹⁰ = 1,024 = 512*(1,024)⁴ - 32*(1,024)² + 1 = 512*(1,000 + 24)⁴ - 32*(1,000 + 24)² + 1 = 512*(1,000,000 + 48,576)² - 32*(1,000,000 + 48,576) + 1 = 512*(1,000,000,000,000 + 99,511,627,776) - 32*(1,000,000 + 48,576) + 1 = 512*1,000,000,000,000 + 512*99,511,627,776) - 32*1,000,000 - 32*48,576 + 1 = 512,000,000,000,000 + (500 + 12)*99,511,627,776 - 32,000,000 - 32*48,576 + 1 = 512,000,000,000,000 + 500*99,511,627,776 + 12*99,511,627,776 - 32,000,000 - 32*48,576 + 1 = 512,000,000,000,000 + 49,755,813,888,000 + (10 + 2)*99,511,627,776 - 32,000,000 - (30 +2)*48,576 + 1 = 512,000,000,000,000 + 49,755,813,888,000 + 10*99,511,627,776 + 2*99,511,627,776 - 32,000,000 - 30*48,576 - 2*48,576 + 1 = 512,000,000,000,000 + 49,755,813,888,000 + 99,511,627,7760 + 199,023,255,552 - 32,000,000 - 1,457,280 - 97,152 + 1 = 562,949,919,866,881

❤ x^+3x = 10^10 +99998 x^2+3x = 100000^2 + 99998 x^2+3x = (y+1)^2 + (y - 1) where y = 99999 x^2+3 x = y^2+3 y (x - y) (x+y+3) = 0 x = y , - y -3 x = 99999 , - 100002

Error 11596

2^10^10^1+2^3/4^4^4^3+2^2^2^2+2^3 1^2^5^2^5^1+1^1^1/2^2^2^2^3+1^1^1^1+1^1^1 1^1^1^1/1^1^1^2^3 /2^3 (x ➖ 3x+2).

562949919866881=2^49-2^25+1

(2^101+8)/(16^13+4^14+8)=562949919866881=2^49-2^25+1 Input (2^101 + 8)/(16^13 + 4^14 + 8) = 562949919866881 Result True Logarithmic form log(4503599895805960, 2^101 + 8) - log(4503599895805960, 16^13 + 4^14 + 8) = log(4503599895805960, 562949919866881)

We know that x^4+4 = (x^2+2x+2)(x^2 - 2x+2) hence Numerator = 2(2^50+2^26+2)(2^50 - 2^26+2) = 2(4^25+4^13+2)(4^25 - 4^13+2) Denominator = 4^26+4^14+8 = 4(4^25+4^13+2) Ans is (4^25 - 4,^13+2)/2 = 2^49 - 2^25 +1

n=x+9 => (n-2)⁴+(n-1)⁴+n⁴+(n+1)⁴ +(n+2)⁴=99 => (n⁴-8n³+24n²-32n+16) +(n⁴-4n³+6n²-4n+1)+n⁴+(n⁴+4n³+6n²+4n+1) +(n⁴+8n³+24n²+32n+16)=99 =>2(n⁴+24n²+16)+2(n⁴+6n²+1)+n⁴ =99 => 5n⁴+60n²+34=99 => 5n⁴+60n²=65 => 5(n⁴+12n²)=5*13 => n⁴+12n²=13 => n²(n²+12)=1*13 =>n=±1 (Ergo: x= -8 or x= -10) Also, n⁴+12n²=13 1n⁴+0n³+12n²+0n-13=0 1 0 12 0 -13 |-1 1 -1 13 -13 0 n³-n²+13n-13=0 1. -1. 13. -13. |1 1. 0. 13. 0 n²+13=0 n=±√13 (Ergo: x= -9+√13 or x= -9-√13) Sic. {x= -8 or x= -10 or x= -9+√13 or x= -9-√13}

{(1+2^49)^2 - 2^50}/ 1+2^49+2^25=1+2^49-2^25 soln.

x = - 2 is a valid solution

1^1^2^2^2^2+2^2^2^2+1^2^2 1^1^1^1+1^1^1^1+1^2 1^2 (x ➖ 2x+1).

(x ➖ 7x+7)^2^2+(x ➖ 4x+2)^2^2+(x ➖ 3x+3)^2^2+(x ➖ 5x+2)^2^2+(x➖ x+11)^2^2 (x ➖ 1x+1)^1^1+(x ➖1 x+1)(x ➖ 2x+2)^1^1+(x ➖ 1x+1)^1^1(x ➖ +1x+1)(x ➖ 5x+5)^1^1(x ➖ 1x+1)^1^1 (x ➖ 2x+1) (x ➖ 1x+1) (x ➖ 2)(x+1) (x ➖ 2x+1).

guess and check "method" (-1)⁴ + (0)⁴ + (1⁴) + (2)⁴ + (3)⁴ = 99 so *x = -8* is a solution so *x = -10* is also a solution [ (-3)⁴ + (-2)⁴ + (-1⁴) + (0)⁴ + (1)⁴ = 99 ] now let's try another way u = x + 9 (u - 2)⁴ + (u - 1)⁴ + u⁴ + (u + 1)⁴ + (u + 2)⁴ = 99 5u⁴ + 2(24u² + 16) + 2(6u² + 1) = 99 5u⁴ + 60u² - 65 = 0 u⁴ + 12u² - 13 = 0 (u² + 13)(u² - 1) = 0 u = ± 1 ∨ u = ± i√13 u = ± 1 => *x = -8* ∨ *x = -10* u = ± i√13 => *x = -9 + i√13* ∨ *x = -9 - i√13*

(x + 7)⁴ = (x + 7)².(x + 7)² (x + 7)⁴ = (x² + 14x + 49).(x² + 14x + 49) (x + 7)⁴ = x⁴ + 14x³ + 49x² + 14x³ + 196x² + 686x + 49x² + 686x + 2401 (x + 7)⁴ = x⁴ + 28x³ + 294x² + 1372x + 2401 (x + 8)⁴ = (x + 8)².(x + 8)² (x + 8)⁴ = (x² + 16x + 64).(x² + 16x + 64) (x + 8)⁴ = x⁴ + 16x³ + 64x² + 16x³ + 256x² + 1024x + 64x² + 1024x + 4096 (x + 8)⁴ = x⁴ + 32x³ + 384x² + 2048x + 4096 (x + 9)⁴ = (x + 9)².(x + 9)² (x + 9)⁴ = (x² + 18x + 81).(x² + 18x + 81) (x + 9)⁴ = x⁴ + 18x³ + 81x² + 18x³ + 324x² + 1458x + 81x² + 1458x + 6561 (x + 9)⁴ = x⁴ + 36x³ + 486x² + 2916x + 6561 (x + 10)⁴ = (x + 10)².(x + 10)² (x + 10)⁴ = (x² + 20x + 100).(x² + 20x + 100) (x + 10)⁴ = x⁴ + 20x³ + 100x² + 20x³ + 400x² + 2000x + 100x² + 2000x + 10000 (x + 10)⁴ = x⁴ + 40x³ + 600x² + 4000x + 10000 (x + 11)⁴ = (x + 11)².(x + 11)² (x + 11)⁴ = (x² + 22x + 121).(x² + 22x + 121) (x + 11)⁴ = x⁴ + 22x³ + 121x² + 22x³ + 484x² + 2662x + 121x² + 2662x + 14641 (x + 11)⁴ = x⁴ + 44x³ + 726x² + 5324x + 14641 (x + 7)⁴ + (x + 8)⁴ + (x + 9)⁴ + (x + 10)⁴ + (x + 11)⁴ = 99 x⁴ + 28x³ + 294x² + 1372x + 2401 + x⁴ + 32x³ + 384x² + 2048x + 4096 + x⁴ + 36x³ + 486x² + 2916x + 6561 + x⁴ + 40x³ + 600x² + 4000x + 10000 + x⁴ + 44x³ + 726x² + 5324x + 14641 = 99 5x⁴ + 180x³ + 2490x² + 15660x + 37600 = 0 x⁴ + 36x³ + 498x² + 3132x + 7520 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the power 3 Let: x = z - (b/4a) → where: b is the coefficient for x³, in our case: 36 a is the coefficient for x⁴, in our case: 1 x⁴ + 36x³ + 498x² + 3132x + 7520 = 0 → let: x = z - (36/4) → x = z - 9 (z - 9)⁴ + 36.(z - 9)³ + 498.(z - 9)² + 3132.(z - 9) + 7520 = 0 (z - 9)².(z - 9)² + 36.(z - 9)².(z - 9) + 498.(z² - 18z + 81) + 3132z - 28188 + 7520 = 0 (z² - 18z + 81).(z² - 18z + 81) + 36.(z² - 18z + 81).(z - 9) + 498z² - 8964z + 40338 + 3132z - 28188 + 7520 = 0 (z⁴ - 18z³ + 81z² - 18z³ + 324z² - 1458z + 81z² - 1458z + 6561) + 36.(z³ - 9z² - 18z² + 162z + 81z - 729) + 498z² - 8964z + 40338 + 3132z - 28188 + 7520 = 0 (z⁴ - 36z³ + 486z² - 2916z + 6561) + 36.(z³ - 27z² + 243z - 729) + 498z² - 5832z + 19670 = 0 z⁴ - 36z³ + 486z² - 2916z + 6561 + 36z³ - 972z² + 8748z - 26244 + 498z² - 5832z + 19670 = 0 z⁴ + 12z² - 13 = 0 ← we can see that (1) is a solution, so we can factorize (z - 1) (z - 1).(z³ + βz² + λz + 13) = 0 → we expand z⁴ + βz³ + λz² + 13z - z³ - βz² - λz - 13 = 0 → we group z⁴ + z³.(β - 1) + z².(λ - β) + z.(13 - λ) - 13 = 0 → we compare with: z⁴ + 12z² - 13 = 0 For z³ → (β - 1) = 0 → β = 1 For z² → (λ - β) = 12 → λ = 12 + β → λ = 13 For z → (13 - λ) = 0 → λ = 13 (of course) (z - 1).(z³ + βz² + λz + 13) = 0 → where: β = 1 (z - 1).(z³ + z² + λz + 13) = 0 → where: λ = 13 (z - 1).(z³ + z² + 13z + 13) = 0 First case: (z - 1) = 0 → z = 1 → it the obvious root (see above) Second case: (z³ + z² + 13z + 13) = 0 z³ + z² + 13z + 13 = 0 ← we can see that (- 1) is a solution, so we can factorize (z + 1) (z + 1).(z² + 13) = 0 Third case: (z² + 13) z² + 13 = 0 z² = - 13 z² = 13i² z = ± i√13 Summarize: z = ± 1 or z = ± i√13 → recall: x = z - 9 First solution: z = 1 → x = - 8 Second solution: z = - 1 → x = - 10 Thirdirst solution: z = i√13 → x = i√13 - 9 Fourth solution: z = - i√13 → x = - i√13 - 9

2+√2. 2-√2

2+i√10. 2-i√10

(x + 7)⁴ + (x + 8)⁴ + (x + 9)⁴ + (x + 10)⁴ + (x + 11)⁴ = 99 Put, t = x + 9 t⁴ + [(t - 2)⁴ + (t + 2)⁴] + [(t - 1)⁴ + (t + 1)⁴] = 99 t⁴ + [(t⁴ - 4*2t³ + 6*2²t² - 4*2³t + 2⁴) + (t⁴ + 4*2t³ + 6*2²t² + 4*2³t + 2⁴] + [(t⁴ - 4*1*t³ + 6*1²*t² - 4*1³*t + 1⁴) + (t⁴ + 4*1*t³ + 6*1²*t² + 4*1³*t + 1⁴)] = 99 t⁴ + 2t⁴ + 48t² + 32 + 2t⁴ + 12t² + 2 = 99 5t⁴ + 60t² - 65 = 0 t⁴ + 12t² - 13 = 0 (t² + 13)(t² - 1) = 0 (t - 1)(t + 1)(t² + 13) = 0 t - 1 = 0 , t + 1 = 0 , t² + 13 = 0 t - 1 = 0 ⇔ t = 1 ⇔ x + 9 = 1 ⇔ x = - 8 t + 1 = 0 ⇔ t = - 1 ⇔ x + 9 = - 1 ⇔ x = - 10 t² + 13 = 0 ⇔ t = ± √(- 13) = ± √13i ⇔ x + 9 = ± √13i ⇔ x = - 9 ± √13i (x ∈ R) reject ∴ x = - 8 , - 10

Correction Note : At 15:56 (14)^2 = 196 instead of 169. Final Answer is 11596. Apology for the mistake 🙏