Відео

x^6 = 64, taking z = 2e^it gives 64e^i(6t) = 64 e^i(6t) = 1 Then the argument 6t has to be some 2kpi, t=0,2pi/6,4pi/6,6pi/6,8pi/6,10pi/6 gives us our 6 solutions in exponential form. t=0 gives z=2 t=pi/3 gives z=sqrt(3) + i t=2pi/3 gives z=-sqrt(3) + i t=pi gives z=-2 t=4pi/3 gives z=-sqrt(3) - i t=5pi/3 gives z=sqrt(3) - i

2 I spent too much time programming with assembler to be scared of 2s powers

Muito bem

4^{x+x ➖ }+{1+1 ➖ }=4{x^2+2}=4^2x^2 16x^2 8^8x^2 4^4x^2 2^2^2^2x^2 1^1^1^1x^2 1x^2 (x ➖ 2x+1).

Just do ³√x^6 = ³√64 x² = 4 √x² = √4 x = 2

x1= 6 x2= -6

2^2× = 20 2xlog 2 = 2log2 + log 5 ÷ log2 2x = 2 + log(2+3) × = 1 + 1/2log5/

4^x= 4×5 (2^2)^×=(2^2)× 5

x^4= 36×36 x^(2^2) =6^4= 6^(2^2) (x^2)^2 =( 6^2)^2 x^4 = 6^4 x=6 ,

x^3^2 (x ➖ 3x+2). 4^16 4^4^4 2^2^2^2^2^2 1^1^1^1^1^2 1^2 (x ➖ 2x+1).

{7+7}=14 7^7 3^4^3^4 1^2^2^3^2^2 1^1^1^3^1^2 3^2:(y ➖ 3x+2).(1)+(2)=3(y ➖ 2x+1).

(n ➖ 1m+1). (mn ➖1 mn+1).

(5) ➖ (19)=95 (5l ➖( 19^1) 5 ➖ (1^1) 5 ➖ (1) (y ➖ 5x+1) (9)+(10)=19 (9)+(5^5) (3^6)+(2^3^2^3) (1^2^3)+(1^1^1^1) (2^3) (y ➖ 3x+2).

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

Just a suggestion- you can change your thumbnail and correct any typos.

wow, that was unnecessary...as soon as you have x log 3 = log 54 just divide by log 3 there leaving you with x = log 54 / log 3

Log54/Log3

well, 2²⁰-2¹⁶ = 2²⁰-2⁶2¹⁰ = 1Mi - 64 Ki = 1024Ki-64 Ki = 960 Ki = 960 × 1024 = 983 040

Just take log base 3 of both sides, save you a million steps

{16+16+16+16}=64 16^16^16^16^ 8^8^8^8^8^8^8^8 2^3^2^3^2^3^2^3^2^3^2^3^2^3^2^3 1^1^1^1^1^1^1^1^1^1^1^1^1^1^1^1^1^2 1^2(x ➖ 2x+1).

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

1^1+2^3/1^1 ➖ 2^3 2^1/2^1 1^1/2^1 1/2 (x ➖ 2x+1).

{4x+4x ➖}=8x^2 2^3x^2 1^1^1x^2 1x^2x (x ➖ 2x+1).

{1+x+}^e^{1 ➖ x ➖ }={2+x^2}=2x^2 (x ➖ 2x+2).

1^6•1^2^3•2^2^2+2^2^2^2+18^18 1^2^3•1^1^1•1^1^1^1•1^1^1^1+9^9 1^1^1 +3^6^3^6 1^2^3^1^2^3 1^1^2^3 2^3 (x ➖ 3x+2).

Im not expert, but this video could end in less than 1:30. People can already solved easily at 1:20 though 😅 Then (a + b )^2 formula just make it more complicated still nice insight..

3^18 3^2^9 3^23^2 1^2^3^1 2^3 (x ➖ 3x+2).

(49)^2 ➖ (48)^2={2401 ➖2304}=13 6^7 6^3^4 2^3^3^2^2 1^1^3^1^2 32 (x ➖ 3x+2).

x^2 (x ➖ 2x+2).

(6)+(6)==12 6^6 2^3^2^3 2^1^1^3 2^3 (y ➖ 3x+2). (xy ➖ 3xy+2).

Hello and thank you for the solution, albeit there is a simpler way to solve this equation: ((X^1/2) + (X- 12)^1/2) = 6. Multiply the LHS of the equation by ((X^1/2 ) - (X- 12)^1/2) / ((X^1/2 ) - (X- 12)^1/2) which is practically 1. Now you get: ((X-X+12) / ((X^1/2 ) - (X- 12)^1/2) equals 6. Simplify and you get: ((X^1/2 ) - (X- 12)^1/2) = 2. Now Add both the original and the last equations and you get: 2(X^1/2) = 8, or (X^1/2) = 4 OR X = 16.

Meus parabéns pela excelente explanação

Y=√x Y^2 = y^y Y=2 or y=1 X=y^2 = 4 or 1

6^(x)^2 ➖ (2)^2=6^{x^2 ➖ 4}=6^x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖}=6^^{x^1+x^1+x^1+x^1}=6^x^4=24 6^6^6^6 2^3^2^3^2^3^2^3 1^1^1^1^1^1^2^3 2^3 (x ➖ 3x+2).

{625x+625x ➖ }+{624x+624x ➖ }={1250x^2+1248x^2}=2498x^4 1250^1248x^4:1^1^5^10^^1^124^24x^4 5^2^5^12^12^12^12 1^2^1^6^6^6^6x^4 2^2^3^2^3^2^3^2^3x^2^2 1^1^1^1^1^1^1^1^3x^1^2 3x^2 (x ➖ 3x+2).

x^2/36=62^3(x ➖ 3x+2). 6/x•6/x =36x^2=6 2^3 (x ➖ 3x+2).

(2)+2ab+(4)=8 (1)+1ab+(2^2) (1^2 ) (ab ➖ 2ab+1)

Excelente explanação

Bro no algebra was needed, I solved this in like 15-20 seconds with pure multiplication lol

That was wonderful 👍👍

{0+0 ➖ }^2=1^2 (x ➖ 2x+1).

{x+x ➖ }+(x)^2 ➖ (12)^2=x^2+{x^2 ➖ 144}=x^2+{x^0+x^0 ➖ x^0+x^0 ➖ }=x^2+{x^1+x^1}={x^2+x^2}='lx^4 (x ➖ 4x+4).

(x ➖ 1x+1) (x ➖1x+1(x ➖ 2x+2) (x ➖ 5x+5)/(x ➖ 1x+1) (x ➖ 1x+1) (x ➖ 2x+2) (x ➖ 5x+5) (x ➖ 1x+1) (x ➖ 1x+1)/(x ➖ 1x+1) (x ➖ 2x+2) (x ➖ 1x+1) /(x ➖ 2x+1) (x ➖ 1x+1) /(x ➖ 2) (x+1)

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

{x^4+x^4 ➖ }+{4+4 ➖ }={x^8+8}=8x^8 2^3x^2^3 1^1^1x^1^1^2 1x^2 (x ➖ 2x+1).

k^2(1-k) = 80 = (-4)^2(1- -4) = 80

A ridiculously primitive problem, clearly not of Olympic difficulty.

1^1^1+1^1/1^1^1 ➖ 1^1 (x ➖ 1x+1).

(x ➖ 5)^4=(x^4 ➖ 625)={x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖}={x^1+x^1+x^1+x^1 }=x^4 (x ➖ 4x+4) 81;3^27 3^3^3 1^1^3 1^3 (x ➖ 3x+1).

(x ➖ 2x+2) {8x+8x ➖ }=16x^2 8^8x^2 2^3^2^3x^2 1^1^1^1^1^1x^2 1^2 (x ➖ 2x+1).