Відео

Please write the x using the other way

<a + <b + <c = 180° изи

Why not put t=\sqrr(x+5)

Похоже, что в 15 секунд не уложился. Но в 20 точно

You can just use the complex plain for this question, by rotating 5 times you’ll get all of the real and complex solutions

X=1 Reason = if knowledge is power, ignorance is bliss

I've no idea what just happened . Felling bored while having lunch , opened UA-cam, this video appeared as the 1st suggestion, played it and now half of my lunch is getting cold and here I'm trying to figure out what is going on. A UA-cam and lunch season turned out to be a confusing mess😅😅

[ 1 + sqr(5)]/2 = Golden ratio, or Golden number or Golden proportion; which is a mathematical constant; then, we have that ([(Golden ratio)^2]^2)^3

Why can’t you just do X^5 = 1 X^5 = cis (0) X = cis (72degrees*k)

Buenas noches Señores: Higher Mathematics. Reciban un cordial saludo. Gracias por este ejercicio y la forma novedosa de darle solución, ¿ cuáles son las soluciones..?. sería bueno que se haga un refuerzo del procedimiento con más ejercicios. agradezco por la atención que se me brinde. Éxitos.

Or you can raise both sides to the 1/√x power and then multiply both sides by -1/2, then apply W function.

졸라말많네

Should state that n and m are POSITIVE integers. Otherwise, the hypothesis of positive results is incorrect.

a=3 first look😂

x = 1.715620733276

Como ingeniero simplemente lo gráfico y veo dónde x hace que la expresion sea igual a 30. Itero para hallar la presicion requerida.

a=3

beautiful

Not simple. I believe that for a not bad school the assignment is for the third grade+ -

British people over complicate sooo much lol. Just use trigonometric form and write x=cos[(2kpi/5)]+isin[(2kpi/5)] where where k is a natural number from 0 to 4

Dude canrt u just do x=1

X = 1 ez

Could it be even more long-winded ?

I got 3+(3*sqrt(6))/3

why dont u just say x = 1 lol because if x = 1 x^5 = 1x1x1x1x1 = 1 1 - 1 = 0 ??

Can't we just use Euler's formula, instead solving the fundamental algebra?

pretty moronic math video with the name Cambridge on it answer is 1

Why he doesn't use demoaver?

e^2nπi/5 for n = 0 to 4 is immediate.

2:42 Instead of computing (x^2)^3 = (x+1)^3 the long way, could you just use the binomial expansion for (x+1)^3 directly?

Just taking your 2nd root (x2) and actually doing what you failed to do, which is to check the validity of your answer by taking that complex and imaginary number to the fifth power, you come up with ca. 3.17623*i. Perhaps this is so in an Alice in the Looking Glass World, or perhaps in a 4th dimensional world where on of the axes is imaginary numbers but how this could possibly be 1 in the REAL world escapes me. What a pointless exercise on your part this has been.

ab = 100 bc = 200 ca = 300 ca = 300 = 3(100) ca = 3(ab) c = 3b bc = 200 b(3b) = 200 3b² = 200 b² = 200/3 b = ±√(200/3) = ±10√2/√3 = ±10√6/3 ab = 100 a(±10√6/3) = 100 a = ±100/(10√6/3) a = ±30/√6 = ±5√6 c = 3b = 3(±10√6/3) c = ±10√6 Test: totals are all positive, so either both multiplicands are positive or both are negative. ab = 100 5√6(10√6/3) = 100 50(6)/3 = 100 300/3 = 100 ✓ bc = 200 (-10√6/3)(-10√6) = 200 100(6)/3 = 200 600/3 = 200 ✓ ca = 300 10√6(5√6) = 300 50(6) = 300 ✓ Solutions: (5√6, 10√6/3, 10√6) (-5√6, -10√6/3, -10√6) a + b + c = 5√6 + 10√6/3 + 10√6 a + b + c = 15√6/3 + 10√6/3 + 30√6/3 [ a + b + c = ±55√6/3 ]

Why is that difficult ??

Waste of time and fucking data!!!

The final solution = 238

1

(sqrt2-1^10 = (2 - 2sqrt2 + 1)^5 = (3 - 2sqrt2)(9 - 12sqrt2 + 8)^2 = (3 -2sqrt2)(289 - 408sqrt2 + 288) = (3 -2sqrt2)(577 - 408sqrt2) = (1731+1632-1224sqrt2-1554sqrt2) = (3363 - 2778sqrt2)

3 три в кубе + три в квадрате =27+9= 36

I won't lie, 80% of high school students in Vietnam can solve this problem easily. it will take a little time but it is still easy

I don’t know if my method is also right but I got a different answer though. Let’s see Let y = sqrt(x) yln(x) = ln(3) (1/2)yln(x) = (1/2)ln(3) yln(y) = ln[sqrt(3)] e^(yln(y)) = e^(ln[sqrt(3)]) ye^y = sqrt(3) W(ye^y) = W(sqrt(3)) y = W(sqrt(3)) But y = sqrt(x) sqrt(x) = W(sqrt(3)) (sqrt(x))^2 = [ W(sqrt(3)) ]^2 x = [ W(sqrt(3)) ]^2 Is it?😶

Well that was a journey

This problem is not testing intelligence. It’s a stupid problem if you know complex algebra. I don’t believe this is a problem in Cambridge entrance exam because Cambridge has course teaching this subject.

ab = 100 bc = 200 ca = 300 c/a = 2 c/b = 3 a/b = 3/2 (3/2)b^2 = 100 so b = sqrt(200/3) 3b^2 = 200 so b = sqrt(200/3) 2a^2 * 300 so a = sqrt(150)=5sqrt6 Check ab = 100 c = 300/sqrt(150) so c = 2sqrt(150) so c= 10sqrt6 Check bc = 10sqrt(200/3)sqrt6 = 200

Mathematics are a pyramid scheme where they have to keep making up stuff to keep working

In one of my math subjects when i was studying degree, i need to prove 1-1=0. Or 1-0=1. Things like that. Really wtf 🙄😶

Its 0 guys, = 0 is 0. /s

8 + 3 = 11.

а=с/2; а=√600/2; с=√600; в=200/√600. Сумма даст 110/√6.

❤❤❤❤

Perfect solution 🎉