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When we know a+b, అండ్ ab Wecan find a-b (a+b)^2- 4ab

If you multiply all the exponents you'd get 3^9/4 /3^9/4 which is 1

Root 27 is root 3 cubed, which is also 3 times root 3. Sub the latter in for the root 27 in the numerator and the former in for the root 27 in the denominator, and you’ll see that the numerator and denominator are equal. Therefore the answer is 1.

(sqrt(3) ^ sqrt(27)) / (sqrt(27) ^ sqrt(3)) (3^(1/2) ^ (3^3)^(1/2)) / (3^3)^(1/2) ^ 3^(1/2)) 3^(1/2 * (3^3)^(1/2)) / 3^(3 * 1/2 * 3^(1/2)) 3^(3^1,5 / 2) / 3^(3^1,5 / 2) = 1 Answer: 1

Why did I get only one value of x √x+3 + √x+3 i = 4 X+3 - (x+3) + 2 (x+3) i =16 X+3 =8/i X=8/i-3 X=(8-3i)/i X=(8i+3)/-1 X=-3-8i

Pretty simple

= 1

√25=5 u video no sense bro

Excellent and nice thanks sir.

What the name of app that record the pen animated?

치환을 적절히 사용하면 더 쉽게 풀 수 있어요

Am i missing something? Why bother with substitution? sqrt(x+3) + sqrt(-x-3) = 4 sqrt(x+3) + i * sqrt(x+3) = 4 sqrt(x+3) * (1 + i) = 4 sqrt(x+3) = 4 /(1+i) sqrt(x+3) = (4/(1+i)) * ((1-i)/(1-i)) = 2 - 2i x+3 = ± sqr(2-2i) = ± (-8i) x= -3 ± 8i

Résultat = 271

And terrible music.

square root of t and square root of negative t. one of them not defined so useless unless imaginary solution required.

Amei

Thank You!

I like the substitution. I just went ahead and started by squaring both sides, simplifying and squaring both sides again, getting a quadratic to solve. But it got fiddly and I made lots of careless mistakes. So the substitution's nice.

Amei!

Repetition is the mother of learning

9(9ª) - (9ª)/9 = 20 y = 9ª 9y - y/9 = 20 (multiply by 9) 81y - y = 180 80y = 180 y = 180/80 y = 9/4 9ª = 9/4 a = ⁹log(9/4) a = ⁹log(9) - ⁹log(4) a = 1 - ⁹log(4) a = 1 - ³log(2)

Если знаешь степени двойки, решается за пару секунд в уме (я помню до 2¹⁸, поэтому чуть дольше 😁) 2²² = 2¹⁸ • 2⁴ = 262144 • 16 = 2 621 440 + 1 572 864 = 4 194 304 4 194 304 - 1 = 4 194 303

will the lamber w function work here?

Thanks prof

n=ln(3/2)/ln(9) , test 9^(ln(3/2)/ln(9)+1)-9^(ln(3/2)/ln(9)-1)=20.25-0.25 , 20.25-0.25=20 , same , OK ,

Прямо любовалась

10:04 Nice solution 👌

ab=100, bc=200 ca=300. a=300/c b=200/c ..... ab= 300/c * 200/c =60000/c2.... 60000/c2=100 c=10 √6 b*10 √6=200 b=20/ √6 a*20/ √6 =100 a=5/ √6

The exponentiation is not associative: usually when we write in the canonical up- indexed way a^b^c we conventionally intend (a^b)^c= a^(bc) and not a^(b^c) ; the same convention extends for multiple exponents if wrote without paranthesis:a^b^c^d= ((a^b)^c)^d= (a^(bc)) ^d=(a^(bcd)) and so on. Thus in this case we should obtain the square root of 2^(540)=2^270; In the other hand, this solution uses the opposite convention, i.e. square root of 2^(6^(2^(1^(5^9))))=square root of 2^(6^(2^(1)))= square root of 2^(6^2)=square root of 2^36=2^18

Х=1

Use binomial theorem, it will be more easy nd short.

b = 100/a; c = 300/a; 100/a * 300/a = 200; a² = 150; a = √150; a * b = 100; √150 * b = 100; b = 100/√150; c * a = 300; c * √150 = 300; c = 300/√150; a + b + c = √150 + 100/√150 + 300/√150 = 150/√150 + 100/√150 + 300/√150 = 550/√150 = 550/5√6 = 110/√6

I'm sorry but there is NO WAY that this is a math Olympiad question !!! I solved it using basic algebra rules in about 23 seconds... CLICKBAIT NONSENSE>>>!

buy a stylus please!!!

NICE TRICK, Thanks

X = 1. Doesn't seem hard. Did it in my head. 😊

3^x (6^x-3^x)/(6^x-3^x)=3 3^x=3 X=1; as simple as this

18^x-9^x=3^x(6^x-3^x); (6^x-3^x) cancels with itself, because exactly the same expression is in the denominator; or 6^x-3^x=t 18^x-9^x/t 18^x-9^x=3^x(6^x-3^x) 3^x * t / t = 3^x =3; x=1

But how 2 is equal to square root 5.may be possible.

It is unnecessarily made complicated. This question doesn’t need to write down. You got 3^x=3 by taking 3*6=18 and 3*3=9.

Isn't this a one-liner? 18^x - 9^x = (3^x) (6^x - 3^x); cancel (6^x - 3^x) top and bottom, leaving 3^x = 3, so => x=1

Х = 1

(18^x - 9^x) / (6^x - 3^x) = 3 ------------------------------------------------------------------------------------------------------- 9^x(2^x - 1) / 3^x(2^x-1)) factor out 9^x on top and 3^x on bot 9^x / 3^x cancel out (2^x-1)/(2^x-1)) = 1 (9/3)^x property: a^x/b^x -> (a/b)^x 3^x = 3 9/3 = 3 x = 1 property: a^b = a -> a=1 done in 1 minute with no music

Find 1/x which will be log2/3log6, similarly find 1/y which will be log3/3log 6. Then add 1/x and 1/y, that will be (x+y/xy) = log2+log3/3log6, on simplifying 1/3. Now take a reciprocal of both sides, you get the answer 3.

9^x(2^x-1)=3*3^x(2^x-1) , 9^x=3*3^x , 3^2x=3*3^x , -> 3^x=3 , 3=3^1 , x=1 , test , x=1 , (18-9)/(6-3)=9/3 , 9/3=3 , same , OK ,

Ответ, а семь и это спектр цветов.

vʼ2+1≈2.414

2^18

I used to solve these kinds of equation in 8th grade😮

Очень простая задача