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√2^8=16; a^{a+1}=2^3=8, so a=2.

I kindly request you: Please learn how to draw an »x« correctly!

sqrt(x)+sqrt(y)=sqrt(80) =4sqrt(5) =[k+(4-k)]sqrt(5) for any 0=<k<4 Hence x=k×sqrt(5), y=(4-k)sqrt(5) for integer k such that 0<k<4

x= +root 2 or -root 2 ?

Was half expecting you to be lazy and just go v-160 =/= v-80.

1/(a+b) +1/(b+c) + 1/(c+a)=7/10 > (a+b+c)/(a+b)+(a+b+c)/(b+c) + (a+b+c)/(c+a) = (a+b+c)*7/10 =7*7/10=49/10 >a/(b+c) +b/(c+a)+ c/(a+b) +3 =49/10 >a/(b+c) + b/(c+a)+ c/(a+b) =49/10 -3=19/10

Why is 80 = 16.5?

why is it so difficult, it can be made much simpler

((1+i)^2):2

For x = 0 => y = 80 => (x , y) = (0 , 80) solution. By symmetry (x , y) = (80 , 0) is another solution.

the video is a bit boring, the song is good at 2x speed

don't forget: sqrt(x/y) = (sqrt(x)/sqrt(y)) only if both x and y are positive (if you're working with real numbers)

Love these vids. Gave me insight

very cool

Checking the results obtained is important since squaring both sides can introduce extraneous roots.

A+3 = {a/(b+c) +1} + {b/(c+a) +1} + {c/(a+b) +1} = (a+b+c){1/(b+c) + 1/(c+a) + 1/(a+b)} = 7*7/10 -> A = 19/10

Before viewing: 900 = 30^2. Then, because (a^m)^n = a^(mn), we can rewrite that as 30^6 Now we want to raise both sides to the fifth power. That way, on the left we will have (x^(x^5))^5 = (x^5)^(x^5), and on the right we have (30^6)^5 = 30^30. Then (x^5)^(x^5) = 30^30. Therefore, x^5 = 30. Then X = 30^(1/5).

This Problem. Thank You.

❤❤❤❤

2=27-25

This question is hard for me,but easy for Soviet olimpiads :) and olimpiad students

(x + y) - xy = 1 => xy = (x + y) - 1 (x + y)² - xy = 7 x + y = b xy = b - 1 b - 1 = b² - 7 b² - b - 6 = 0 (b - 3)(b + 2) = 0 x + y = b = 3 xy = 2 t² - 3t + 2 = 0 t = (3 ± 1)/2 => t = 2 ∨ t = 1 x + y = b = -2 xy = -3 t² + 2t - 3 = 0 t = (-2 ± 4)/2 => t = 1 ∨ t = -3 *(x, y) = {(1, 2); (2, 1); (1, -3); (-3, 1)}*

How did you onow you had to do that to solve the problem? I'm not that good at maths....

I promise you math is gonna divide humanity. No pun intended

7!=5040

3^x+(3^x)²=(3^x)³ (3^x)[(3^x)²-(3^x)-1]=0 x=0 or 3^x=ß, golden ratio =½[1±sqrt(5)] x=[log(ß)]/log(3)

If 2^x=5, then x=log(2,5) by definition.

x = √(4 + √7) + √(4 - √7) , x > 0 x² = (4 + √7) + (4 - √7) + 2*√(16 - 7) = 8 + 6 = 14 x = ±√14 { Discard x ≤ 0 } x = √14 ≈ 3.7416573867739...

Let y = 3ˣ . For x ∈ ℝ , y > 0 and x = log(y)/log(3) . y + y² = y³ y³ - y² - y = 0 y*(y² - y - 1) = 0 y = 0, (1 ± √5)/2 { Discard y ≤ 0 } y = (√5 + 1)/2 x = log((√5 + 1)/2)/log(3) ≈ 0.4380178794859...

Lack an approximate result in a normal digits. Without any counting and calculating I assessed x as ca 0,7, in 10 sec.

Both x and y are the complex cube roots of -8.

MUSIC IS ANNOYING !

X=(-1)

But why does the average in the 891*899 works in this solution?

After you move -1 to the left side, you can factor the equation by grouping.

ძალიან უკულტუროდ წერენ, აჯღაბაჯღად.

26^3

(x-1)(x^5+….+1=0,x^6-1=0,x=e^iπn/3,n=(1,2,3,4,5)

move one to the other side, use geometric series sum, the answer is clearly then the 6th roots of unity excluding 1

There's no way this is an Olympiad problem.

3*9^x = 99 x = log9(33) Done

cmmonnn dont be afrain to hitem with an instant log base 9 to not have to worry about it. sloppy answers are correct answers!

could have solve after the first factoring .

i used another method: 9^x= 3^2x let y = 3^x we the, get y^2+y^2+y^2=99 y^2+y^2+y^2-99=0 3y^2 - 99=0 using the quadritic formula y= +(root(4*3*99))/6 or y= -(root(4*3*99))/6 we take the positif solution because we are going to use the lorgarthms 3^x= +(root(4*3*99))/6 x=ln((squareroot(4*3*99))/6) / ln(3) x=1.59132916932

Great❤

4 or -2 and as a jee aspirant did mentally

X(X-(2/X)-1)=0=X^2-2-X=(X-2)(X+1) -> X=2 or -1 ans: 2+(2/2)+1=4 or -1+(2/-1)+1=-2

26³ 27³ Nice trick ❤

= 20^10 * 40^-5 = 2^10 * 10^10 * 4^-5 * 10^-5 = 2^10 * 10^10 * (2^2)^-5 * 10^-5 = 2^10 * 2^-10 * 10^10 * 10^-5 = 2^(10 - 10) * 10^(10 - 5) = 10^5

I'm studying for jee and this took me like 8 seconds