Відео

At the point where a²-(1/2a)=0 We have difference of two squares. Or? Couldn we have used the identity (a-b)(a+b)?

We can notice that x<=2 so inside radical exists √(2-x)<=2 so x>=-2 so outside radical exists X>=0 Also x=1 is a solution to equation 1=√2-√(2-1) Now by squaring x^2=2-√(2-x) √(2-x)=2-x^2 squarig again 2-x=4-4x^2+x^4 X^4-4x^2+x+2=0 (we know x-1 is solution) R x^4. X^3. X^2. X. 1 1. 0. -4. 1 2 1. 1. 1. -3. -2 0 2. 1. 3. 3. 4. Not a root -2. 1. -1. -1 0. root So we have x^2-x-1=0 or x=1 or x=-2 rejected (<0) X3=(1+√5)/2 x4=(1-√5)/2 rejected as it is <0 If we are using x3 then √(2-x)=√(3-√5)/2=√(6-2√5)/4=1/2(√5-1) Then the outer root is √2-(√5-1)/2=√(5-√5)/2 does not satisfy equation

Amazing. I messed up calculations trying some substitutions but intorducing a square was impressive. Still after substituting a and b i woukd not square just use the power of 1/2 as square root getting √x^√y=9=3^2 √y^√x=8=2^3 Wonder if this is sufficient to get the solutions or will itnkead to omission

(2a+1)b=44-a^2 . 2a+1>=3 b=(44-a^2)/(2a+1)=(1/4)*{(-2a+1)+175/(2a+1)} (2a+1)(2a+4b-1)=175 (2a+1,2a+4b-1)=(5, 35),(7, 25)

(2a+1)b=44-a^2>0.a=1,2,3,4,5,6. (a,b)=(2,8),(3,5)

You a pro and an excellent teacher . THANK YOU!!!

a shall > 0

Muito bom. Parabens

5/a + 6/b =7 => 1+6=7 => 5/a=1 , a=5 => 6/b=6 , b=1 => 5+2=7 => 5/a=5 . a=1 => 6/b=2 , b=3

X=4

Essa eu achei difícil

mi solucion solo mirando la miniatura m=-3 ya las otras posibles soluciones las dejo para el video mi primera vista mi opcion fue el 3, pero como luego vi que el m^2 estaba antes del m^3 supuse que era un numero negativo, total m^2 siempre dara como resultado un numero positivo y al m^3 quedar negativo daria como resultado -(-x) lo cual da otro numero positivo y solo queda sumarlos XD

In3^x-2=lnx then (x-2)ln3=1lnx then x-2=1 and 3=x then x=3

Why do you calculate again? If a+1=5, then a=4, isn't it?

3 &4

(4 + √15)ˣ + (4 - √15)ˣ = 62 { (4 - √15)*(4 + √15) = 1 } [(4 + √15)ˣ]² - 62*(4 + √15)ˣ + 1 = 0 (4 + √15)ˣ = 31 ± √(31²-1) = 31 ± 8*√15 x = ±2

問題にｘは実数とは書いていない。だから虚数解も解としなければならない。

무지하게 복잡하게 해설

If doing the other stuff is not necessary, then using log to base 4 would have minimized the steps needed to solve the equation as log properties mean applying log to the appropriate base on both sides of the equation means on one side, the base and its respective log would cancel out, leaving whatever is on the exponent as another log property already drops the exponent to the front of the log and multiplied. In this case, log to base 4 will result in solving the equation with minimal steps assuming log of 24 to base 4 is it without doing anything else.

√20 √√√ root 2 π theory #

√（6－√６－ｘ）＝ｘ 6－√（6－ⅹ）＝ⅹ＾２ 6－ⅹ^2=√（6ーｘ） 36ー12x＾２＋ｘ^4-＝6-x ｘ^4-12x^2+x= -30 x(x^3-12x+1)=-30 if x=1 x^3-12x+1=10 1*(10) = 10 if x=2 x^3-12x+1=- -15 2*(-15) = -30 OK! if x=3 x^3-12x+1=- -8 3*(-8) = -24 if x=4 x^3-12x+1=-17 4*(17) =136 if x=5 x^3-12x+1=-66 2*(66) =330 X=2

a²+2ab+b²=44+b²-b Only b=5 and b=8 give perfect square. Then calculate a.

Math Olympiad: a + b = 6(√ab); a/b =? a + b + 2(√ab) = 8(√ab), (√a + √b)² = 8(√ab), √a + √b = ± 2(√2)(⁴√ab) a + b - 2√(ab) = 4(√ab), (√a - √b)² = 4(√ab), √a - √b = ± 2(⁴√ab) 2√a = ± 2(√2)(⁴√ab) ± 2(⁴√ab) = ± 2(√2 ± 1)(⁴√ab), √a = ± (√2 ± 1)(⁴√ab) 2√b = ± 2(√2)(⁴√ab) -/+ 2(⁴√ab) = ± 2(√2 -/+ 1)(⁴√ab), √b = ± (√2 -/+ 1)(⁴√ab) (√a)/(√b) = √(a/b) = (√2 ± 1)/(√2 -/+ 1) = [(√2 ± 1)(√2 ± 1)]/[(√2 -/+ 1)(√2 ± 1)] = [(√2 ± 1)²]/(2 - 1) = (3 ± 2√2), a/b = (3 ± 2√2)² = 9 ± 12√2 + 8 = 17 ± 12√2

x^4-12x^2-x+30=0 x=-2 16-48+2+30=0

3:38, not 4, but 2😊

(-a)^1/a=4 , (-1/a)^-1/a=2^2 , -1/a=2 then a=-1/2

у=44/х, х+44/х-20=0. *х х²-20х+44=0. D=400-44*4=400-176=300-76=270-6=264. √D=2√66 x=(20±22√66)/2. x₁=10+11√66⇒y₁=20-10-11√66=10-11√66⇒(x₁;y₁)=(10+11√66;10-11√66); x₂=10-11√66⇒y₂=20-10+11√66=10+11√66⇒(x₂;y₂)=(10-11√66;10+11√66).

4^x=24 log(4^x)=log(24) x×log(4)=log(24) x=log(24)÷log(4) x=2.29248125...

خودتو نمودی و حلش نکردی

The answer is incorrect

rearrange the eq to: b = [44-a^2] / [2a + 1]. since both a, b > 0 a <= 6. Only options a =2,3 give integer solutions for b (8,5). This might be an easier solve.

2,8 just by putting a=2 and solving for b, if b comes to be a positive integer, its the correct answer

44 is small. Just try A from 1 til 6.

nice

Very easy

x + y = 20 ← this is the sum S xy = 44 ← this is the product P x & y are the solution of the following equation: a² - Sa + P = 0 a² - 20a + 44 = 0 Δ = (- 20)² - (4 * 44) = 400 - 176 = 224 = 2 * 112 = 2 * 2 * 56 = 2 * 2 * 4 * 14 = 16 * 14 a = (20 ± 4√14)/2 a = 10 ± 2√14 First solution: x = 10 + 2√14 y = 10 - 2√14 Second solution: x = 10 - 2√14 y = 10 + 2√14

You can simplify a bit from 7:07. y^2 - y + 1 - (y + 1)(y - 1)^2 = 0 y^2 - (y - 1) - (y - 1)^2*(y + 1) = 0 y^2 - (y - 1)[1 + (y - 1)(y + 1)] = 0 y^2 - (y - 1)[1 + y^2 - y + y - 1] = 0 y^2 - (y - 1)(y^2) = 0 y^2(1 - [y - 1]) = 0 y^2(1 - y + 1) = 0 y^2(2 - y) = 0 y^2 = 0 or 2 - y = 0 y = +/-sqrt(0) or 2 - y + y = 0 + y y = 0 or 2 = y y = 0 or y = 2 y^3 = 0^3 or y^3 = 2^3 y^3 = 0 or y^3 = 8 Since x = y^3, x = 0 or x = 8 Only 8 is a solution to sqrt(1 + sqrt[1+x]) = cbrt(x).

8 is the only solution. If you plug -1 or 0 into the equation, you'll find they are not solutions.

Not an olympiad question at all.

x=(_/x)^(_/x) Domain: x[>=]0 Let n=_/x, n²=x n²=nⁿ ln(n²)=ln(nⁿ) 2[ln(n)]=n[ln(n)] n[ln(n)]-2[ln(n)]=0 [ln(n)](n-2)=0 ln(n)=0 n=e⁰ n=1 _/x=1 x=1 ❤ n-2=0 n=2 _/x=2 x=4 ❤

x=4

Plug 2 you get sqart of 6 - sqart sqart of 6+2 = sqart 6 - sqart sqart 8 = sqart 6-2=2

why it cannt be (1,44/3)

Слишком просто. А если произведение x и y равнялось 41 вместо предложенного 44? Сделайте нестандартное решение.

22 & -2 But quest may be wrong.

Hi I said that because when I tried on my tablet calculator didn’t match the answer. But now I will certainly try on a different system.

Obviously x^2 >= 5 let y^2= x+5 then x^2 - 5=y y^2-x = x^2-y x^2-y^2+x-y = 0 (x-y) (x+y+1) = 0 (x^2-x-5)(x^2+x-4) = 0 now we will use quadratic formula and condition x^2 >= 5 to get roots.

It's a classic. There's a trick here. Square it up. you will get a quadratic equation with respect to 5 (five) with the parameter X. Solve it. The discriminant will be an exact square (2X+1). After that, solve the resulting 4 quadratic equations.

X=1

4a^2+8ab+4b=176 4a^2-1+4b(2a+1)=175 (2a-1)(2a+1)+4b(2a+1)=175