4

Also, if x² = -11, you should have two roots: + ✓11i and -✓11i

It's wrong! If x = 0, you should have 14 / (-14) = -1

Very nice! Well done and essy to follow!

When you get to 2^8 why break it up to 2^4 x 2^4. Know your powers of 2's. 2^8 =256.

Nice

Nice

u did nothing

Option A

70

16

16

4

B

625

X = 0.5

A

The numbers are perfectly tractable and dont need a calculator. 125+512+1331-1320=648=12x54

It’s is somewhat simpler and can be solved mentally: I) You see 4 square roots, hence it’s power of 1/16; ii) To get to 16 you have to have 2 in power of 4; iii) you multiple 16 by 4 to get 64

What is the logic behind dividing by 3 on both sides

??? The quotient in the problem was 8 yet it mysteriously changed to 18 when working out the solution??

Nice

Ok but is no one gonna talk about how they over complicate it by making the numbers sqrt(x^2)

This sh*t cannot be a math olympiad question. Just stop fooling people into wasting time on fake olympiad questions.

b = -(x1 + x2) c = x1*x2 Vieta theorem

For those interested, I the four answers are: x=2, x=-11, x=(-9+√(-159))/2, x=(-9-√(-159))/2. I found the two real solutions using Gauss method.

So good solve

Other solution : Easy to see that we have to multiply 4 successive integers, so we expect the result to be more or less a power of 4 of the integers. It is then easy to test some integers. 6^4 = 36*36 = 1296, we are almost there ! Then we take the 4 numbers as slightly above a 6*6*6*6, let's try 5*6*7*8. It works :)

Practical solution I found. (x+6)!/(x+2)! = (x+6)(x+5)(x+4)(x+3). you work with factorials, the answer is expected to be integer. Decompose. 1680 = 2*2*2*2*3*5*7. Reorganise as a product of 4 following integers, with obviously 7 in them (you wont have one of the numbers begin 14 or 21), you find quicly 1680 = 5*6*7*8. x+3=5, so x=2.

Легко решить если учесть что 11-sqrt120=(sqrt 6-sqrt 5)^2 Тогда sqrt (sqrt 5+sqrt 6)*sqrt (sqrt 6-sqrt 5)=sqrt ((sqrt 6-sqrt 5)*(sqrt 6?sqrt 5))=sqrt (6-5)=1 30 секунд, а вы решаете 5 минут

After watching the video: Yes, you can do it the complicated way, but... Just a little bit thinking outside formulas would have been easier...

Well... This translates to: Find the 4 consecutive numbers, whose product is 1680, and substract 2 of the smallest to get X... Easy...

Una demostración clara de este problema.

Nicely done! I did this one a bit differently: I decided to do A=(2^333 * 111^333)/(3^222 * 111^222) and prove it is >1 It easily simplifies to ((8^111)/(9^111)) * 111^111 which I rearranged as ((11^111)/(9^111)) * 8^111. Now 11/9 > 1 => (11^111)/(9^111) > 1 and 8^111 > 1 => A > 1 and thus 222^333 > 333^222 It is quite similar in the end, but I think I prefer your technique :)

One can guess that 222^333 must be a lot bigger because the exponent has a much bigger effect than the base. To make this rigorous, note that 333^222 < 1000^222 = 10^666. On the other hand, 222^333 > 100^333 = 10^666.

Nice

9

Why does the x look like a n?

why u complicating this 1/0.5 is 2 so solve from there

At step 5, where we say n^3 + n^2 - (3^3 + 3^2) = 0, why can't we move the (3^3 + 3^2) across the = to show n^3 + n^2 = 3^3 + 3^2? At that point, isn't it obvious that n=3?

Interesting

9

23

23

😊😊

9

you are making it complicated when it can be solved easily😂 5x/ √0.25 = 200. √0.25 = 0.5 200 x 0.5 = 5x 100 = 5x X = 100/5 = 20

13

13

Well evaluated