Відео

E=9-4(5)^(1/2)

This is a Golden Ratio problem. Divide the numerator and denominator by √7: [(√5-1)/2]^6=? (√5-1)/2 is the positive root of φ^2+φ-1=0 so φ^2=1-φ; φ^4=1-2φ+φ^2=2-3φ; and φ^6=(2-3φ)(1-φ)=2-5φ+3φ^2=5-8φ Thus, ?=5-8(√5-1)/2=9-4√5

(5^7 ➖7/7^4)^6 (5^^1 ➖ 1/1^4)^3^2.(5^1/2^2)^3^2 (1^1/1^1)^3^2 ()^3^2 (x ➖ 3x+2).

111^1 (y ➖ 111x+1). 111^1 (x ➖ 111y+1).

Good job👍

(3)+4)=7 (x ➖ 4x+3) (x ➖ 3x+3).

Good morning,l think can be by squaring both sides then by taking all terms one side and solving equation by taking common,we get x=5

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By intuition, I thought exactly the opposite. I was completely wrong...

This is not correct since sqrt(x) and x^(1/2) is not the same. You must be very careful since sqrt(x) and sqrt4(x) always has a positive real part. For x=3 the result of the expression to the left is 81 which is the same as the number to the right but for x=-3 the result of the expression is 81*i and for x=3*i or x=-3i the result to the left is -81.

nice solution 👌👌

gli esercizi proposti sono scritti in maniera chiara e precisa

This is a very nice sum.😊😊😊

(x ➖ 3)^2/21+(x ➖ 2)^2/22+(x ➖ 1)^2/23=(x^2 ➖ 9)/21+(x^2 ➖ 2)/22+{x^2 ➖ 1}=23 ={x^0+x^0 ➖ }/21+{x^0+x^0 ➖ }/22+{x^0+x^0 ➖ }/23={x^1/21+x^1/22+x^1/23}= x^3/66 x^3/3^2^3^2 x^1/3^1^1^2 /32 (x ➖ 3x+2).

For those interested in solving for a and b: a = 1/729 and b = 177147.

This is a simple "sight" problem - x = 24

After log property apply componando and dividendo. Solve

thanks for sharing

Nice problem, just please note, that thumbnail shows this problem: 3^(x+y) = 25 3^(x-y) = 125 While you are solving this problem: 3^(x-y) = 25 3^(x+y) = 125 Which results in different answers, 1/243 vs 243.

Your thumbnail image has a different equation, which I evaluate here: x + 1/x = 1 { Note that x ≠ 0 } x² - x + 1 = 0 { Multiply by x } x = (1 ± i*√3)/2 { Quadratic formula } = cis(±π/3) = cis(±60⁰) I notice that they're the complex conjugate pair of the 3 roots for x³ = -1 . That is, x³ + 1 = (x + 1)*(x² - x + 1) = 0 , but without x = -1 . Anyway, for x²⁰²⁵ , each x³ is -1 , so we break down 2025 into the nearest multiple of 3 , plus an offset of 0 or ±1 . Well, 2025 = 675*3 + 0 , so x²⁰²⁵ = (x³)⁶⁷⁵ = (-1)⁶⁷⁵ = -1 . Thus, we find: x²⁰²⁵ + 1/x²⁰²⁵ = -1 + 1/-1 = -2 Your video has a different equation, which I evaluate here: x + 1/x = √2 { Note that x ≠ 0 } x² - √2*x + 1 = 0 { Multiply by x } x = (√2 ± i*√2)/2 { Quadratic formula } = (1 ± i)*√2/2 = cis(±π/4) = cis(±45⁰) I notice that they're a complex conjugate pair of the 8 roots of x⁸ = 1 . Also, we know that, for n ∈ ℤ , xⁿ = cis(±n*π/4) = cis(n*45⁰) . Thus, we see: x²⁰²⁵ = cis(±2025*π/4) = cis(±2025*45⁰) We know it cycles every 8 times by 2*π or 360⁰, so we take 2025 mod 8 to get a remainder of 0..7 to use as the equivalent power of x. Well, 2025 mod 8 = 1 , so we have: x²⁰²⁵ = cis(±π/4) = cis(±45⁰) = x Thus, we find: x²⁰²⁵ + 1/x²⁰²⁵ = x + 1/x = √2 We can do a generalized version. Find: xⁿ + 1/xⁿ , n x ∈ ℕ1 . Given: x + 1/x = k , k ∈ ℝ . We know that x ≠ 0 . x² - k*x + 1 = 0 { Multiply x and rearrange } x = (k ± √(k² - 4))/2 { Quadratic formula } For k² ≤ 4 , the 2 roots form a complex conjugate pair, and can be written as: x = (k ± i*√(4 - k²))/2 , -2 ≤ k ≤ 2 For now, we'll restrict k to -2 ≤ k ≤ 2 . In polar form, we write them as r∠θ . Magnitude r : r = √(k² + 4 - k²)/2 = 1 . Angle θ : θ = ±arccos(k/2) . They land on the unit circle of the complex number plane. Thus, we have: xⁿ = rⁿ∠n*θ , n ∈ ℕ1 = 1ⁿ∠±n*arccos(k/2) = 1∠±n*arccos(k/2) xⁿ = cis(±n*arccos(k/2)) = cis(n*θ) Since each x is the reciprocal of the other x , the property extends to xⁿ and its reciprocal: 1/xⁿ = x⁻ⁿ = cis(∓n*arccos(k/2)) Thus, their sum is a real number -- the real part of x doubles and the imaginary part cancels out: xⁿ + 1/xⁿ = xⁿ + x⁻ⁿ = 2*cos(±n*arccos(k/2)) Also note that, if m*θ is equivalent to an integer multiple of 2*π , for m ∈ ℕ1 and m < n , then xᵐ = 1 and xⁿ = x⁽ⁿ ᵐᵒᵈ ᵐ⁾ . This is a way to reduce the exponent from n to n' = n mod m , if m exists. Thus, xⁿ = cis(±n'*arccos(k/2)) . This also applies to the sum: xⁿ + 1/xⁿ = xⁿ + x⁻ⁿ = 2*cos(±n'*arccos(k/2)) Even if m doesn't exist for large n , the original formula works fine, as the cosine function cycles for every 2*π or 360⁰ of its argument, so its argument may have smaller equivalents. We can test this formula with both cases in the video. Case 1: For k = 1 and n = 2025 , we compute θ first: θ = ±arccos(k/2) = ±arccos(1/2) = ±π/3 or ±60⁰ I notice that 6*θ = ±2*π or ±360⁰ , so m = 6 and n' = n mod m = 2025 mod 6 = 3 . Thus: x²⁰²⁵ = x³ = cis(3*θ) = -1 x²⁰²⁵ + 1/x²⁰²⁵ = x³ + x⁻³ = 2*cos(3*θ) = -2 Even without finding m = 6 , we can still use the original formula, which will have ±2025*60⁰ for the cosine function. That argument is ±(337*360⁰ + 180⁰) and is equivalent to ±180⁰ . They return the same result. Case 2: For k = √2 and n = 2025 , we compute θ first: θ = ±arccos(k/2) = ±arccos(√2/2) = ±π/4 or ±45⁰ I notice that 8*θ = ±2*π or ±360⁰ , so m = 8 and n' = n mod m = 2025 mod 8 = 1 . Thus: x²⁰²⁵ = x¹ = cis(1*θ) = (1 ± i)*√2/2 x²⁰²⁵ + 1/x²⁰²⁵ = x¹ + x⁻¹ = 2*cos(1*θ) = √2 Even without finding m = 8 , we can still use the original formula, which will have ±2025*45⁰ for the cosine function. That argument is ±(253*360⁰ + 45⁰) and is equivalent to ±45⁰ . They return the same result.

x^(2 + 7/4 - -1/4) = 81 x^4 = 3^4 x = 3*cis(k*π/2) , k = 0, ±1, 2 = ±3, ±3*i

5^(n+2)=4ⁿ 25(5ⁿ)=4ⁿ (⅘)ⁿ=25 (⅘)ⁿ=5² n=log_⅘(5²) n=2[log_⅘(5)] ❤

³√(√4)*2³ = 8*³√2 = 2^(10/3) = ³√1024 ≈ 10.079368399158...

((1 + √5)/2)¹² = [[ ((1 + √5)/2)²*((1 + √5)/2) ]²]² = [[ ((6 + 2*√5)/4)*((1 + √5)/2) ]²]² = [[ ((3 + √5)/2)*((1 + √5)/2) ]²]² = [[ (8 + 4*√5)/4 ]²]² = [[ 2 + √5 ]²]² = [ 9 + 4*√5 ]² = 161 + 72*√5 ≈ 321.99689437998...

√(799² + 799 + 800) = √(799² + 799 + 799 + 1) = √(799 + 1)² = | 799 + 1 | = 800

√8^√2 / √2^√8 = √2^(3*√2 - 2√2) = √2^√2 = 2^(√2/2) ≈ 1.6325269194381... BTW, √2/2 = 1/√2 = 2^-(1/2) = 2^-2^-1 , so we can write: x = 2^2^−2^−1 { How cool is that? } This works with the normal precedence rules for exponentiation.

5⁽ˣ⁺²⁾ = 4ˣ (x + 2)*log(10/2) = x*log(2²) (x + 2)*(1 - log(2)) = x*2*log(2) x*(1 - 3*log(2)) = -2*(1 - log(2)) x = -2*(1 - log(2))/(3 - 3*log(2) - 2) x = -1/(3/2 - 1/(1 - log(2))) ≈ −14.42513487802... If you're fond of log base 10 lookup tables and 4-function calculators without parentheses or a memory storage, then you can rewrite x as: x = -1/(1/(log(2) - 1) + 1.5) Then, you can use .30103 as an approximate value for log(2) . I get x ≈ −14.42513672479... using a postfix (RPN) calculator with this sequence: [.30103][Enter^][1][-][1/x][1.5][+][1/x][+/-]

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Nice👌

This is definitely not an Olympiad question😂😂

that spider is proud of himself after he getting into the video

This video was helpful, I really struglle with algebra but I understood your video more than my teacher, keep up the good work man

1:44 😂😂

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Nice👌

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Nice job👌

Nice👌👌

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nice solution