Відео

5

WAY too fiddly. =5 is immediately obvious. There are 3 solutions since it is a cubic, symmetrical about the origin on the Argand plane. so x = 5, 5 cis(2π/3) and 5 cis(-2π/3). giving x = 5, 5(-1+i√3)/2, 5(-1-i√3)/2 And if this is the level I need for Harvard, let me at it!

4제곱x=24 4*x×4*x=24 X*x=24÷16=1.5 X=루트1.5=1.224 4×1.224×4×1.224=24 4.896×4.896=24 X=4.896 X=(4=제곱2).896 4.896 4.896×4.896=24 N에 제곱은? N*x×n*x. =n=2. 3제곱값 구하는방식 찾는데 니들은 껌이지. 5에 x제곱=35 5×x×5×x=35 X*x=35÷25=1.4 루트1.4=1.1832 5×1.1832×5×1.1832=35 ----------------------- =2.1832×2.1832=35 검산해보면 거의틀리는데 검산해보니 이건 맞음❤

Second solution: (k/4)⁶ = 2⁶ k⁶ / 4⁶ = 2⁶ k⁶ = 2⁶ · 4⁶ k⁶ = 8⁶ k⁶ − 8⁶ = 0 (k³ + 8³) ⋅ (k³ − 8³) = 0 (k + 8) ⋅ (k² − 8k + 8²) ⋅ (k − 8) ⋅ (k² + 8k + 8²) = 0 Rule of zero product: 1. k + 8 = 0 k₁ = −8 2. k² − 8k + 8² = 0 k₂,₃ = 4 ± √(4² − 8²) = 4 ± √−48 = 4 ± i · √48 k₂ = 4 − i · 4√3 = 4 · (1 − i · √3) k₃ = 4 + i · 4√3 = 4 · (1 + i · √3) 3. k − 8 = 0 k₄ = 8 4. k² + 8k + 8² = 0 k₅,₆ = −4 ± √((−4)² − 8²) = −4 ± √−48 = −4 ± i · √48 k₅ = −4 − i · 4√3 = 4 · (−1 − i · √3) k₆ = −4 + i · 4√3 = 4 · (−1 + i · √3)

First solution: (k/4)⁶ = 2⁶ k⁶ / 4⁶ = 2⁶ k⁶ = 2⁶ · 4⁶ k⁶ = 8⁶ kₙ = 8 · [cos(n · 360°/6) + i · sin(n · 360°/6)] k₁ = 8 · [cos(60°) + i · sin(60°)] = 4 + i · 4 · √3 = 4 · (1 + i√3) k₂ = 8 · [cos(120°) + i · sin(120°)] = −4 + i · 4 · √3 = 4 · (−1 + i√3) k₃ = 8 · [cos(180°) + i · sin(180°)] = −8 + i · 0 = −8 k₄ = 8 · [cos(240°) + i · sin(240°)] = −4 − i · 4 · √3 = 4 · (−1 − i√3) k₅ = 8 · [cos(300°) + i · sin(300°)] = 4 − i · 4 · √3 = 4 · (1 − i√3) k₆ = 8 · [cos(360°) + i · sin(360°)] = 8 + i · 0 = 8

X=2 2×2×2×2+2×2=16+4=20

Muito bom, consegui resouver esse bicho cabeludo.

K=8e^i*2*n*pi/6 n=0,1,..,n Which is 8*(cos(2n pi/6)+i sin(2n pi/6)) Which includes +\-8 but as well +\-0.5+\-isqrt(3)/2

(5×5)+(5×5)+(5×5)+(5×5)+(5×5)= X*X*X=125 125÷x÷(x*x)= x*x=125÷x X=125÷x÷x X=125÷5=25 X=25÷5=5 어떻한 수+같은수로 두번 나눠줘야 x에 3제곱이다? X에 3제곱정의는 N 이 어떻한수로, !!!어떻한수와 같은 동일한수로 2번!!! 나눠줘야 n 에 3제곱이다=!!!같은수로 3번곱해야 3제곱!!!이다. 125=1×125 5×25 5×5×5 나눌땐 동일한수로 2번 나눠지고, 곱할땐 같은수로 3번 곱하는거"3제곱 8÷2÷2=2 8=2×2×2 N을 동일한 x로 한번더 x =x가 되는게 n에 3제곱 #####10에 x÷x=x 10에x÷x=x는?###### 이걸풀면 필즈상 아벨상 받는다ㅋㅋ

2⁶=±2⁶ K=±8

X^3=5^3 X*x*x= 5*5*5 Y=5

six to the boro nine to the boro

Keep growing my fellow Mathematician ❤❤❤❤❤❤I have subscribed

5

Lk dn ion h bhai

X³=5³? X=5

Это задача не уровня Гарварда. Так, 8-9 класс школы максимум. Основная идея для решения - что корней такого уравнения - три по максимальной степени.

125

basic algebra: a cubic equation has 3 solutions; most comments only mention x=5. there two more solutions. x = [-5 ± √(∆) ]/2 = [-5 ± √(-75) ]/2 = [-5 ± √(25 * -1 * 3) ]/2 = [-5 ± 5*i*√ 3]/2 where (by definition) i = √(-1)

in English ... "a*b" is read "a times b" not "a into b". very distracting.

x=5

Hello, I found 3 solutions : (x,y) : {(148,141),(68,51),(1012,1011)}

Why doesn’t X = 5?

The solution is clearly obvious. Not sure why the creator took the marathon route through the jungle to solve this problem.

25x = 250 x = Log 250 / log25 x = 1.71533 25^1.71533 = 250

3x = 5^3 3x= 125 x = Log125 / Log3 x = 4.39492 3^4.39492 = 125

11^x=11^5=[[ x=5 ACADEMICMARCELIUS MARTIROSIANAS 2o18 VIKIPEDIA Volnaencyklopedia ACADEMIC MARCELIUS Martirosianas 17 Matwematikos ModelioAtradejas Plic Stanford Cardinal universityPNG

(x-2)(x^3+2x^2+5x+10)=0 , x=2 , x^3+2x^2+5x+10=0 , x^2(x+2)+5(x+2)=0 , (x+2)(x^2+5)=0 , x=-2 , x= i*V5 , -i*V5 ,

X³=5³; solution; X³-5³=0; (X-5)(X²+5X+25)=0; X-5=0; X=5; X²+5X+25=0; ∆=5²-4×1×25; ∆=25-100; ∆=-75;S={5}.

IMAGINARIO

X^3=5^3 X=5

That set zero multi zero. Zero multi not zero Not zero multi zeo Empty. And, zeo multi zero is zero necesary the proof Also zero multi not zero is zero proof.