Відео

asnwer=11 isit

Math olympiad level ? I cannot believe it.

x=ln216/ln2 , y=ln216/ln3 , let a=x , and let b=y , a*b/(a+b)= 3 , ln216/ln2 * ln216/ln3 / (ln216/ln2 + ln216/ln3) , ((ln216)^2/(ln2*ln3))/(ln216 *ln3 + ln216*ln2)/(ln2*ln3 ) , ln216/(ln2+ln3) = 3 , OK ,

271

10000 100 +100= 200 81-10=71 271

/100 2 +81 2 +19 4 /2

Inpes tasi pantasan bayakya yang berinpestasidan menam saham karna utk bertaruh bila menang akan mendapatkan ke untungan.

2^22-1; 2^22-2^0; 2^0(2^22-1= 4194303

Почему очень расписываете решение. Красота решения в краткости оформления. Спасибо большое.

3^x=33 , 3^k=33 , k=ln33/ln3 , 3^x=3^k , 3^x=3^(ln33/ln3 ) , -> x=ln33/ln3 , test , 3^(ln33/ln3)=33 , / x=~ 3.18266 , / , OK ,

The final result can be found in the area formula for a regular octagon: A=2*(1+sqrt 2)s^2.

잘보고 있습니다

Такие ДОЛГИЕ ПРЕДВПРИТЕЛЬНЫЕ ЛАСКИ - а КОНЕЦ, всё равно "ОДИН"!

255/16

영상이 참 좋네요

Amei!

i enjoy your videos . I am stuck at home with covid and it is fun to watch your process with this problem . I am a retired math teacher

The longest way to solve IT)

I just solved it using a function and found the intersection with y At 5 x

Its ok, but last should be catch a2+b2= x, and ab = y then formula apply, I think it will be easy.

1+√5/2 =phi

Здорово!

Também é possível encontrar os valores de a, b e c escrevendo na forma de matriz e fazendo o escalonamento.

[x+y*sqrt(2)]^3=…=(x^3+6*x*y^2)+(3*y*x^2+2y^3)*sqrt(2)=99+70*sqrt(2) .Therefore (1) x*(x^2+6*y^2)=99 ; (2) y*(3*x^2+2*y^2)=70 . (3) x=3*u (4) y=2*v .We substituted (3) in (1) end (4) in (2) . We get : (5) u*(3*u^2+8*v^2)=11 ; (6) v*(27*u^2+8*v^2)=35 . You can guess, that : !!! u=v=1 !!! We get : !!!! x=3 , y=2 !!!! Therefore : [3+2*sqrt(2)]^3=99+70*sqrt(2) . 3+2*sqrt(2)=[1+sqrt(2)]^2 . We get your answer !!! With respect , Lidiy

Typical case of intersection between a circle and a line. Every time there are zero, one or two solution points.

Awsome!

HVALA

Решала так же. И ответ тот же. Только для того, чтоб понять, что (1+2**1/2)**3=7+5*2**1/2, что неочевидно, решала систему х**3+6*х*у**2=7 3*х**2*у+2*у**3=5, где искомое число=х+у**2**1/2

Dünndvhiss

√[125+50√6]=√[a²+2ab+b²]=√(a+b)²=a+b a²+b²=125 2ab=50√6 ab=25√6 25√6=√[2*3*5*5*5*5]=√75*√50 a²=75 a=√75=5√3 b²=50 b=√50=5√2 Answer : 5√3 + 5√2

Its 1(rootx) + i(rootx)=4 (1+i)(rootx)=4 Rootx =4/(1+i) x=[4/(1+i)]^2 x=16/2i x=8/i??

9:30 ...hmm... sqr(x^2) = |x| = ±x

√(16√8) = 4[2^(3/4)] = 2^(2+3/4) = 2^(11/4) = 64^x = 2^6x 6x = 11/4; x = 11/24

This can be done so much faster. Just square immediately: x + 2✓x✓-x -x = 16 ✓x✓-x = 8 Squaring again gives -x^2 = 64 Which gives x = +- 8i

1,0,2

√a+√b=10 √ab=10 a+b+2√ab=100 a+b=80 a+b==100-20=80 (a+b)² -4ab=(a-b) ² =80²-400 ( 80+20) (80-20) =6000 6000=2×3×5×2×10×10 a-b=20√15 a+b=80 2a=80+20√15=20(4+√15) a=10(4+√15) b=80-10(4+√15)

√[16 * √8] = 64^(x) √[16 * √2^(3)] = 64^(x) √[16 * {2^(3)}^(1/2)] = 64^(x) √[16 * 2^(3/2)] = 64^(x) √[2^(4) * 2^(3/2)] = 64^(x) √[2^(11/2)] = 64^(x) [2^(11/2)]^(1/2) = 64^(x) 2^(11/4) = 64^(x) 2^(11/4) = [2^(6)]^(x) 2^(11/4) = 2^(6x) 11/4 = 6x x = 11/24

√2³⁶ = 2¹⁸ = 2¹⁶ · 4 = 65.536 · 4 = 262.144

5 минут возни - и двойки наконец сократились. А в конце пришлось вспоминать, чему равны a и b. Если бы заранее не записали, не решили бы.

again, like many problems in this category: YOU HAVE TO STATE WHAT IS THE SET THAT YOU ARE LOOKING FOR SOLUTIONS (in this case is R and not C, deducted after the presented solution)

я просто понял сначала что число под корнем не может быть отрицательным и поэтому полюбому x должен быть 0

sqrt(16*sqrt(8))=64^x (16*8^(1/2))^(1/2)=(2^6)^x (2^4*2^(3/2))^1/2)=2^(6*x) (2^(4+3/2))^(1/2)=2^(6*x) (2^(11/2))^(1/2)=2^(6*x) 2^((11/2)*(1/2))=2^(6*x) 2^(11/4)=2^(6*x) 11/4=6*x x=(11/4)/6 x=11/24

Bazi durumlari iki kere yazmak sacma zaman kaybi bunlari çözerken hizli olmak gerekli

or just... use the quadratic formula

What a waste of time sind ink !

You made a mistake. Multiplication is denoted solely by a (), not with the additional central dot.

There are many similar problems, so many people will solve it easier. If you look at the "x²+x" part, you can expect that the equation is {x+(A+1)}(x-A)=x²+{(A+1)-A}x-A(A+1)=x²+x-A(A+1)=0 (A: positive integer). In other words, since A(A+1)=44442222, you just need to find two integers that differ by 1. 44442222 is 44442222=(2222)(20001). Here, adding up the digits of 20001 gives 2+0+0+0+1=3, so that it is a multiple of 3, and 20001=(3)(6667). Therefore, 44442222=(2222)(3)(6667)=(6666)(6667), as a result, we can find that A is 6666. (x+6667)(x-6666)=0 ⇒x=-6667, 6666.

Δε κατάλαβα πως λύνεται

Observation: -44442222<0, meaning the roots differ in sign. Observation: (x-r)(x+s)=0 will result in roots r and -s. In this problem, the coefficient of the x term is positive, so s+(-r) > 0, or s>r. Note the inverted relationship: s+(-r)>0, yet the roots are +r and -s. Once you have 44442222 = 6a(6a+1) , you can substitute a=1111, yielding 6666(6667). 6667 > 6666, making the roots r=6666 , s=-6667 .

X^2+X=44442222 X=6666 X=-6667